Maharashtra Board Class 11 Physics Chapter 5 Gravitation Solutions

Choose The Correct Option.

Question 1. The value of acceleration due to gravity is maximum at
(A) the equator of the Earth.
(B) the centre of the Earth.
(C) the pole of the Earth.
(D) slightly above the surface of the Earth.
Answer:
(C) the pole of the Earth.
In simple words: The Earth's poles are closer to its center than the equator, and gravitational acceleration is inversely proportional to the square of the distance from the center, making it maximum at the poles.

🎯 Exam Tip: Remember that Earth's shape (oblate spheroid) causes gravity to vary, being highest at the poles and lowest at the equator.

Question 2. The weight of a particle at the centre of the Earth is ______.
(A) infinite.
(B) zero.
(C) same as that at other places.
(D) greater than at the poles.
Answer:
(B) zero.
In simple words: At the Earth's center, the gravitational forces from all directions cancel out, resulting in a net gravitational force of zero, hence zero weight.

🎯 Exam Tip: Understanding the distribution of mass and its effect on gravitational force is key to solving problems about gravity inside celestial bodies.

Question 3. The gravitational potential due to the Earth is minimum at
(A) the centre of the Earth.
(B) the surface of the Earth.
(C) a points inside the Earth but not at its centre.
(D) infinite distance.
Answer:
(A) the centre of the Earth.
In simple words: Gravitational potential is minimum (most negative) at the center of the Earth because the maximum amount of work is required to move a mass from there to infinity.

🎯 Exam Tip: Gravitational potential is a scalar quantity and is always negative; its minimum value refers to the most negative value, indicating maximum binding.

Question 4. The binding energy of a satellite revolving around planet in a circular orbit is 3 x 10⁹ J. Its kinetic energy is _________
(A) 6 × 10⁹ J
(B) -3 × 10⁹ J
(C) -6 × 10⁺⁹ J
(D) 3 × 10⁺⁹ J
Answer:
(D) 3 × 10⁺⁹ J
In simple words: For a satellite in a circular orbit, the kinetic energy is numerically equal to the binding energy. If binding energy is 3 x 10⁹ J, then kinetic energy is also 3 x 10⁹ J.

🎯 Exam Tip: For circular orbits, remember the relationship: Total Energy (E) = -Kinetic Energy (K) = (1/2)Potential Energy (U). Binding energy is -E, so Binding Energy = K.

Answer The Following Questions.

Question 1. State Kepler's law equal of area.
Answer:
The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
In simple words: A planet moves faster when it's closer to the Sun and slower when it's farther away, but the area covered by the line connecting it to the Sun remains constant over equal time periods.

🎯 Exam Tip: Kepler's second law, also known as the law of areas, is a direct consequence of the conservation of angular momentum.

Question 2. State Kepler's law of period.
Answer:
The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.
In simple words: The further a planet is from the Sun, the longer it takes to complete one orbit, and there's a specific mathematical relationship between its orbital period and its average distance.

🎯 Exam Tip: Kepler's third law, \(T^2 \propto r^3\), is crucial for understanding orbital mechanics and calculating orbital periods or radii.

Question 3. What are the dimensions of the universal gravitational constant?
Answer:
The dimensions of universal gravitational constant are: \([L^3M^{-1}T^{-2}]\).
In simple words: The universal gravitational constant 'G' has dimensions that combine length, mass, and time, derived from Newton's law of gravitation (F = GMm/r²).

🎯 Exam Tip: To find the dimensions of a constant, rearrange its defining formula and substitute the dimensions of known quantities (Force: \(MLT^{-2}\), Mass: \(M\), Length: \(L\)).

Question 4. Define binding energy of a satellite.
Answer:
The minimum energy required by a satellite to escape from Earth 's gravitational influence is the binding energy of the satellite.
In simple words: Binding energy is the amount of energy needed to completely free a satellite from the Earth's gravity, making its total energy zero or positive.

🎯 Exam Tip: Binding energy is always positive and equal in magnitude to the absolute value of the satellite's total mechanical energy in orbit.

Question 5. What do you mean by geostationary satellite?
Answer:
Some satellites that revolve around the Earth in equatorial plane have same sense of rotation as that of the Earth. The also have the same period of rotation as that of the Earth i.e.. 24 hours. Due to this, these satellites appear stationary from the Earth's surface and are known as geostationary satellites.
In simple words: A geostationary satellite is one that orbits the Earth directly above the equator, taking exactly 24 hours to complete an orbit, which makes it appear motionless from the ground.

🎯 Exam Tip: Key characteristics of a geostationary satellite include orbiting in the equatorial plane, a 24-hour period, and appearing fixed relative to Earth's surface, making them ideal for communication.

Question 6. State Newton's law of gravitation.
Answer:
Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
In simple words: Any two objects with mass attract each other; the stronger the attraction, the more massive the objects, and the weaker it is the farther apart they are.

🎯 Exam Tip: Remember the mathematical form \(F = G \frac{m_1 m_2}{r^2}\), and note that the force is always attractive and acts along the line joining the centers of the two masses.

Question 7. Define escape velocity of a satellite.
Answer:
The minimum velocity with which a body should be thrown vertically upwards from the surface of the Earth so that it escapes the Earth 's gravitational field, is called the escape velocity (\(v_e\)) of the body.
In simple words: Escape velocity is the slowest speed an object needs to be launched from a planet's surface to completely break free from its gravitational pull without further propulsion.

🎯 Exam Tip: Escape velocity depends only on the mass and radius of the planet from which the object is escaping, not on the mass or direction of the object itself.

Question 8. What is the variation in acceleration due to gravity with altitude?
Answer:
Variation in acceleration due to gravity due to altitude is given by, \(g_h = g \left(\frac{R}{R+h}\right)^2\)
where,
\(g_h\) = acceleration due to gravity of an object placed at h altitude
\(g\) = acceleration due to gravity on surface of the Earth
\(R\) = radius of the Earth
\(h\) = altitude height of the object from the surface of the Earth.
Hence, acceleration due to gravity decreases with increase in altitude.
In simple words: As you go higher above the Earth's surface, the acceleration due to gravity decreases because you are further from the Earth's center, reducing the gravitational pull.

🎯 Exam Tip: The decrease in 'g' with altitude is proportional to the inverse square of the distance from the Earth's center, making it a significant factor for satellites and high-altitude experiments.

Question 9. On which factors does the escape speed of a body from the surface of Earth depend?
Answer:
The escape speed depends only on the mass and radius of the planet.
[Note: Escape velocity does not depend upon the mass of the body]
In simple words: The speed needed to escape a planet's gravity relies solely on the planet's mass and its size, not on the mass of the object trying to escape.

🎯 Exam Tip: The formula \(v_e = \sqrt{\frac{2GM}{R}}\) clearly shows dependence only on the planet's mass (M) and radius (R), and the gravitational constant (G).

Question 10. As we go from one planet to another planet, how will the mass and weight of a body change?
Answer:
1. As we go from one planet to another, mass of a body remains unaffected.
2. However, due to change in mass and radius of planet, acceleration due to gravity acting on the body changes as, \(g \propto \frac{M}{R^2}\).
Hence, weight of the body also changes as, \(W \propto \frac{M}{R^2}\)
In simple words: Your mass stays the same no matter which planet you're on, but your weight changes because each planet has a different mass and radius, leading to a different gravitational pull.

🎯 Exam Tip: Distinguish clearly between mass (an intrinsic property, constant) and weight (a force, varies with local gravity). Weight is \(W = mg\).

Question 11. What is periodic time of a geostationary satellite?
Answer:
The periodic time of a geostationary satellite is same as that of the Earth i.e., one day or 24 hours.
In simple words: A geostationary satellite takes exactly one Earth day, or 24 hours, to complete one orbit, matching Earth's rotation period.

🎯 Exam Tip: The 24-hour orbital period is a defining characteristic of geostationary satellites, allowing them to remain above the same point on the Earth's equator.

Question 12. State Newton's law of gravitation and express it in vector form.
Answer:
1. Statement:
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
2. In vector form, it can be expressed as,
\( \vec{F}_{21} = G \frac{m_1 m_2}{r^2} (-\hat{r}_{21}) \)
where, \(\hat{r}_{21}\) is the unit vector from \(m_1\) to \(m_2\).
The force \(\vec{F}_{21}\) is directed from \(m_2\) to \(m_1\).
In simple words: Newton's law of gravitation states that any two masses attract each other, with the force depending on their masses and the distance between them. In vector form, it shows that this attractive force acts along the line connecting the two masses.

🎯 Exam Tip: The negative sign in the vector form \((-\hat{r}_{21})\) indicates that the force is attractive, pointing opposite to the direction of the unit vector from \(m_1\) to \(m_2\).

Question 13. What do you mean by gravitational constant? State its SI units.
Answer:
1. From Newton's law of gravitation,
\(F = G \frac{m_1 m_2}{r^2}\)
where, G = constant called universal gravitational constant Its value is \(6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2\).
2. \(G = \frac{Fr^2}{m_1 m_2}\)
If \(m_1 = m_2 = 1 \text{ kg}\), \(r = 1 \text{ m}\) then \(F = G\).
Hence, the universal gravitational constant is the force of gravitation between two particles of unit mass separated by unit distance.
3. Unit: \(\text{N m}^2/\text{kg}^2\) in SI system.
In simple words: The gravitational constant 'G' is a fundamental number that quantifies the strength of the gravitational force between any two masses, specifically, it's the force between two 1-kilogram objects placed 1 meter apart.

🎯 Exam Tip: The value of G is very small, reflecting the weakness of gravity compared to other fundamental forces, and its unit \(\text{N m}^2/\text{kg}^2\) is derived directly from Newton's law of gravitation.

Question 14. Why is a minimum two stage rocket necessary for launching of a satellite?
Answer:
1. For the projection of an artificial satellite, it is necessary for the satellite to have a certain velocity.
2. In a single stage rocket, when the fuel in first stage of rocket is ignited on the surface of the Earth, it raises the satellite vertically.
3. The velocity of projection of satellite normal to the surface of the Earth is the vertical velocity.
4. If this vertical velocity is less than the escape velocity (\(v_e\)), the satellite returns to the Earth's surface. While, if the vertical velocity is greater than or equal to the escape velocity, the satellite will escape from Earth's gravitational influence and go to infinity.
5. Hence, minimum two stage rocket, one to raise the satellite to desired height and another to provide required horizontal velocity, is necessary for launching of a satellite.
In simple words: A two-stage rocket is needed to launch a satellite because the first stage lifts it to a specific altitude, and the second stage then provides the crucial horizontal velocity required to enter orbit, preventing it from either falling back to Earth or escaping entirely.

🎯 Exam Tip: Successful satellite launch requires both achieving the correct altitude and, more importantly, imparting a precise horizontal (orbital) velocity, which a multi-stage rocket effectively manages by shedding mass as fuel is consumed.

Question 15. State the conditions for various possible orbits of a satellite depending upon the horizontal speed of projection
Answer:
The path of the satellite depends upon the value of horizontal speed of projection \(v_h\) relative to critical velocity \(v_c\) and escape velocity \(v_e\).
Case (I) \(v_h < v_c\):
The orbit of satellite is an ellipse with point of projection as apogee and Earth at one of the foci. During this elliptical path, if the satellite passes through the Earth's atmosphere. it experiences a nonconservative force of air resistance. As a result it loses energy and spirals down to the Earth.
Case (II) \(v_h = v_c\):
The satellite moves in a stable circular orbit around the Earth.
Case (III) \(v_c < v_h < v_e\):
The satellite moves in an elliptical orbit round the Earth with the point of projection as perigee.
Case (IV) \(v_h = v_e\)
The satellite travels along parabolic path and never returns to the point of projection. Its speed will be zero at infinity.
Case (V) \(v_h > v_e\):
The satellite escapes from gravitational influence of Earth traversing a hyperbolic path.
In simple words: A satellite's orbit depends on its initial horizontal speed: if too slow, it crashes; if just right, it orbits circularly; if faster but still below escape velocity, it orbits elliptically; at escape velocity, it leaves on a parabolic path; and above escape velocity, it leaves on a hyperbolic path.

🎯 Exam Tip: Memorize the critical velocity (\(v_c\)) and escape velocity (\(v_e\)) benchmarks, as they define the boundaries for circular, elliptical, parabolic, and hyperbolic trajectories.

Answer The Following Questions In Detail.

Question 1. Derive an expression for critical velocity of a satellite.
Answer:
Expression for critical velocity:
1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
2. If the satellite is moving in a circular orbit of radius \((R + h) = r\), its speed must be equal to the magnitude of critical velocity \(v_c\).
3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
\( \therefore \) Centripetal force = Gravitational force
\( \frac{mv_c^2}{r} = \frac{GMm}{r^2} \)

\( \implies v_c^2 = \frac{GM}{r} \)

\( \implies v_c = \sqrt{\frac{GM}{r}} \)

\( \therefore v_c = \sqrt{\frac{GM}{(R+h)}} \)
Using \(g_h = \frac{GM}{(R+h)^2}\), so \(GM = g_h (R+h)^2\)
\( \implies v_c = \sqrt{\frac{g_h (R+h)^2}{(R+h)}} \)

\( \implies v_c = \sqrt{g_h (R+h)} \)
This is the expression for critical speed at the orbit of radius \((R + h)\).
In simple words: To find the critical velocity of a satellite, we balance the centripetal force required for circular motion with the gravitational force acting on it. This leads to a formula that depends on the Earth's mass, the radius of the orbit, and the gravitational constant.

🎯 Exam Tip: Clearly show the equilibrium between gravitational and centripetal forces. The final expression \(v_c = \sqrt{\frac{GM}{R+h}}\) or \(v_c = \sqrt{g_h(R+h)}\) are key for evaluating this derivation.

Question 2. State any four applications of a communication satellite.
Answer:
Applications of communication satellite:
1. For the transmission of television and radiowave signals over large areas of Earth's surface.
2. For broadcasting telecommunication.
3. For military purposes.
4. For navigation surveillance.
In simple words: Communication satellites are essential for transmitting TV and radio signals, enabling global phone calls and internet, assisting military operations, and providing accurate navigation services like GPS.

🎯 Exam Tip: Focus on how communication satellites facilitate long-distance data transmission, broadcasting, and global positioning, highlighting their widespread utility.

Question 3. Show that acceleration due to gravity at height h above the Earth's surface is \(g_h = g \left(\frac{R}{R+h}\right)^2\)
Answer:
Variation of g due to altitude:
1. Let,
R = radius of the Earth,
M = mass of the Earth.
g = acceleration due to gravity at the surface of the Earth.
2. Consider a body of mass m on the surface of the Earth. The acceleration due to gravity on the Earth's surface is given by,
\(g = \frac{GM}{R^2}\) ....(1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी को एक गोले के रूप में दर्शाता है जिसका केंद्र O और त्रिज्या R है। पृथ्वी की सतह पर द्रव्यमान m दर्शाया गया है। यह आरेख गुरुत्वाकर्षण के कारण त्वरण 'g' को पृथ्वी की सतह पर दर्शाता है।
3. The body is taken at height h above the surface of the Earth as shown in figure. The acceleration due to gravity now changes to,
\(g_h = \frac{GM}{(R+h)^2}\) ....(2)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी को केंद्र O और त्रिज्या R के साथ दर्शाता है। एक वस्तु जिसका द्रव्यमान m है, पृथ्वी की सतह से h ऊँचाई पर बिंदु P पर स्थित है। यह आरेख गुरुत्वाकर्षण के कारण त्वरण 'gh' को ऊँचाई 'h' पर दर्शाता है।
4. Dividing equation (2) by equation (1), we get,
\( \frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}} \)

\( \therefore \frac{g_h}{g} = \frac{R^2}{(R+h)^2} \)

\( \therefore g_h = g \frac{R^2}{(R+h)^2} \)
We can rewrite,
\( \therefore g_h = g \left(\frac{R}{R+h}\right)^2 \)
\( \therefore g_h = g \left[R \cdot \frac{1}{R(1 + \frac{h}{R})}\right]^2 \)
\( \therefore g_h = g \left[ \frac{1}{(1 + \frac{h}{R})} \right]^2 \)
\( \therefore g_h = g \left(1 + \frac{h}{R}\right)^{-2} \)
Using binomial expansion \((1+x)^n \approx 1+nx\) for small \(x\), where \(x = h/R\) and \(n = -2\):
\( \text{v.} \) For small altitude h, i.e., for \(\frac{h}{R} \ll 1\), by neglecting higher power terms of \(\frac{h}{R}\),
\( g_h = g \left(1 - \frac{2h}{R}\right) \)
This expression can be used to calculate the value of g at height h above the surface of the Earth as long as \(h \ll R\).
In simple words: The acceleration due to gravity decreases with altitude because the distance from the Earth's center increases. By comparing the gravitational acceleration at the surface with that at height 'h', we derive a formula showing this inverse square relationship.

🎯 Exam Tip: The derivation requires understanding Newton's law of gravitation and algebraic manipulation. Pay attention to the conditions under which the binomial approximation \((1 + h/R)^{-2} \approx 1 - 2h/R\) is valid.

Question 4. Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक उपग्रह के विभिन्न प्रक्षेपवक्र को दर्शाता है जो उसके स्पर्शरेखीय प्रक्षेपण गति पर निर्भर करता है। इसमें पृथ्वी के चारों ओर एक वृत्त (जब \(v_h = v_c\)), दो दीर्घवृत्त (एक अपोजी के रूप में प्रक्षेपण बिंदु के साथ जब \(v_c < v_h < v_e\) और दूसरा जब \(v_h < v_c\)), एक परवलय (जब \(v_h = v_e\)), और एक अतिपरवलय (जब \(v_h > v_e\)) पथ दिखाए गए हैं, जो गुरुत्वाकर्षण से बचने के लिए आवश्यक गति को दर्शाता है।
\(v_h\) = horizontal speed of projection
\(v_c\) = critical velocity
\(v_e\) = escape velocity
In simple words: This diagram illustrates how the path a satellite takes around Earth changes based on its initial horizontal speed: too slow, it falls; just right, it orbits circularly; faster, it orbits elliptically; at escape velocity, it leaves on a parabolic path; and even faster, it leaves on a hyperbolic path.

🎯 Exam Tip: Accurately drawing and labeling the different trajectories (circular, elliptical, parabolic, hyperbolic) with respect to \(v_c\) and \(v_e\) is crucial for full marks in such a question.

Question 5. Derive an expression for binding energy of a body at rest on the Earth's surface.
Answer:
1. Let,
M = mass of the Earth
m = mass of the satellite
R = radius of the Earth.
2. Since the satellite is at rest on the Earth, \(v = 0\)
\( \therefore \) Kinetic energy of satellite.
\(K.E = \frac{1}{2} mv^2 = 0\)
3. Gravitational potential at the Earth's surface
\( = - \frac{GM}{R}\)
\( \therefore \) Potential energy of satellite = Gravitational potential \( \times \) mass of satellite
\( = - \frac{GMm}{R}\)
4. Total energy of satellite = T.E = P.E + K.E
\( \therefore \text{ T.E.} = - \frac{GMm}{R} + 0 = - \frac{GMm}{R}\)
5. Negative sign in the energy indicates that the satellite is bound to the Earth, due to gravitational force of attraction.
6. For the satellite to be free form Earth's gravitational influence, its total energy should become positive. That energy is the binding energy of the satellite at rest on the surface of the Earth.
\( \therefore \text{ B.E.} = \frac{GMm}{R}\)
In simple words: The binding energy of an object at rest on Earth's surface is the amount of positive energy needed to overcome its negative total energy (gravitational potential energy) and completely free it from Earth's gravity.

🎯 Exam Tip: The key steps are recognizing that kinetic energy is zero for a body at rest and that binding energy is the positive value of the negative total energy (which is equal to the negative of potential energy when kinetic energy is zero).

Question 6. Why do astronauts in an orbiting satellite have a feeling of weightlessness?
Answer:
1. For an astronaut, in a satellite, the net force towards the centre of the Earth will always be, \(F = mg - N\).
where, N is the normal reaction.
2. In the case of a revolving satellite, the satellite is performing a circular motion. The acceleration for this motion is centripetal, which is provided by the gravitational acceleration g at the location of the satellite.
3. In this case, the downward acceleration, \(a_d = g\), or the satellite (along with the astronaut) is in the state of free fall.
4. Thus, the net force acting on astronaut will be, \(F = mg - mg\) i.e., the apparent weight will be zero, giving the feeling of total weightlessness.
In simple words: Astronauts feel weightless in orbit because both they and the satellite are constantly falling around the Earth together. Since there's no supporting force pushing up against them, they experience an apparent lack of weight.

🎯 Exam Tip: The crucial concept here is "free fall." Astronauts are not experiencing zero gravity; rather, they are continuously falling towards Earth but constantly missing it due to their high horizontal velocity, hence no normal force is exerted on them.

Question 7. Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth's surface.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख पृथ्वी के केंद्र से दूरी के साथ गुरुत्वाकर्षण के कारण त्वरण 'g' में परिवर्तन को दर्शाता है। ग्राफ दिखाता है कि केंद्र से सतह तक 'g' रैखिक रूप से बढ़ता है, अधिकतम मान पर पहुंचता है, और फिर सतह से दूर ऊँचाई के साथ घटता है।
In simple words: The graph shows that acceleration due to gravity increases linearly from zero at the Earth's center to a maximum at the surface, and then decreases as you move further away from the surface (with increasing altitude).

🎯 Exam Tip: Ensure your graph accurately shows a linear increase of 'g' from the center to the surface and an inverse square law decrease beyond the surface. Label axes (g and r or distance) clearly.

Question 8. At which place on the Earth's surface is the gravitational acceleration maximum? Why?
Answer:
1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
2. At poles, latitude \(\theta = 90^\circ\).
\( \therefore g' = g\)
i.e., there is no reduction in acceleration due to gravity at poles.
3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The polar radius of the Earth is 6356 km which minimum. As \(g \propto \frac{1}{R^2}\), acceleration due to gravity is maximum at poles i.e., \(9.8322 \text{ m/s}^2\).
In simple words: Gravitational acceleration is highest at the Earth's poles because the poles are slightly closer to the Earth's center than the equator, and there's no reduction due to the Earth's rotation at the poles.

🎯 Exam Tip: Remember the two main reasons for maximum gravity at poles: smaller radius (due to Earth's oblate spheroid shape) and zero effect of centrifugal force from Earth's rotation.

Question 9. At which place on the Earth surface the gravitational acceleration minimum? Why?
Answer:
1. Gravitational acceleration on the surface of the Earth depends on latitude of the place as well as rotation and shape of the Earth.
2. At equator, latitude \(\theta = 0^\circ\).
\( \therefore g' = g - R\omega^2\)
i.e., the acceleration due to gravity is reduced by amount \(R\omega^2 (\approx 0.034 \text{ m/s}^2)\) at equator.
3. Also, shape of the Earth is actually an ellipsoid, bulged at equator. The equatorial radius of the Earth is 6378 km, which is maximum. As \(g \propto \frac{1}{R^2}\) acceleration due to gravity is minimum on equator i.e., \(9.7804 \text{ m/s}^2\).
In simple words: Gravitational acceleration is lowest at the Earth's equator because the equator is farther from the Earth's center and the Earth's rotation causes a slight outward centrifugal force, reducing the effective gravity.

🎯 Exam Tip: The two key reasons for minimum gravity at the equator are the larger radius (due to Earth's bulge) and the maximum effect of the outward centrifugal force due to Earth's rotation.

Question 10. Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain altitude.
Answer:
The minimum energy required by a satellite to escape from Earth 's gravitational influence is the binding energy of the satellite.
Expression for binding energy of satellite revolving in circular orbit round the Earth:
1. Consider a satellite of mass m revolving at height h above the surface of the Earth in a circular orbit. It possesses potential energy as well as kinetic energy.
2. Let M be the mass of the Earth, R be the Radius of the Earth, \(v_c\) be critical velocity of satellite, \(r = (R + h)\) be the radius of the orbit.
3. Kinetic energy of satellite = \(\frac{1}{2}mv_c^2 = \frac{1}{2} \frac{GMm}{r}\)
4. The gravitational potential at a distance r from the centre of the Earth is \( - \frac{GM}{r}\)
\( \therefore \) Potential energy of satellite = Gravitational potential \( \times \) mass of satellite
\( = - \frac{GMm}{r}\)
5. The total energy of satellite is given as T.E. = K.E. + P.E.
\( = \frac{1}{2} \frac{GMm}{r} - \frac{GMm}{r} = - \frac{1}{2} \frac{GMm}{r}\)
6. Total energy of a circularly orbiting satellite is negative. Negative sign indicates that the satellite is bound to the Earth, due to gravitational force of attraction. For the satellite to be free from the Earth's gravitational influence its total energy should become zero or positive.
7. Hence the minimum energy to be supplied to unbind the satellite is \(+ \frac{1}{2} \frac{GMm}{r}\). This is the binding energy of a satellite.
In simple words: The binding energy of an orbiting satellite is the positive energy required to liberate it from Earth's gravity. It's equal to the absolute value of the satellite's total mechanical energy, which is half the absolute value of its gravitational potential energy.

🎯 Exam Tip: The derivation involves calculating both kinetic and potential energies of an orbiting satellite. Remember that for a circular orbit, \(KE = \frac{1}{2}|PE|\), and total energy \(TE = KE + PE = -\frac{1}{2}|PE|\). Binding energy is \(-TE\).

Question 11. Obtain the formula for acceleration due to gravity at the depth 'd' below the Earth's surface.
Answer:
1. The Earth can be considered to be a sphere made of large number of concentric uniform spherical shells.
2. When an object is on the surface of the Earth it experiences the gravitational force as if the entire mass of the Earth is concentrated at its centre.
3. The acceleration due to gravity on the surface of the Earth is, \(g = \frac{GM}{R^2}\)
4. Assuming that the density of the Earth is uniform, mass of the Earth is given by
M = volume \(\times\) density \( = \frac{4}{3} \pi R^3 \rho\)
\( \therefore g = \frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2} = \frac{4}{3} \pi R p G\) ....(1)
5. Consider a body at a point P at the depth d below the surface of the Earth as shown in figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र पृथ्वी के एक खंड को दर्शाता है जिसमें केंद्र O और त्रिज्या R है। बिंदु P को पृथ्वी की सतह से गहराई 'd' पर दिखाया गया है, जिसका मतलब है कि यह केंद्र से (R-d) दूरी पर है। यह आरेख यह समझाने के लिए उपयोग किया जाता है कि गहराई के साथ गुरुत्वाकर्षण त्वरण कैसे बदलता है, जहां केवल आंतरिक क्षेत्र (R-d) द्रव्यमान ही बिंदु P पर बल लगाता है।
Here the force on a body at P due to outer spherical shell shown by shaded region, cancel out due to symmetry.
The net force on P is only due to the inner sphere of radius OP = R - d.
6. Acceleration due to gravity because of this sphere is,
\(g_d = \frac{GM'}{(R-d)^2}\)
where,
\(M' = \text{volume of the inner sphere} \times \text{density}\)
\(M' = \frac{4}{3} \pi (R-d)^3 \times \rho\)
\( \therefore g_d = \frac{G \times \frac{4}{3} \pi (R-d)^3 \rho}{(R-d)^2} \)
\( \therefore g_d = G \times \frac{4}{3} \pi (R-d) \rho\) ....(2)
7. Dividing equation (2) by equation (1) we get,
\( \frac{g_d}{g} = \frac{G \times \frac{4}{3} \pi (R-d) \rho}{G \times \frac{4}{3} \pi R \rho} \)
\( \therefore \frac{g_d}{g} = \frac{R-d}{R} \)
\( \therefore g_d = g \left(1 - \frac{d}{R}\right) \)
This equation gives acceleration due to gravity at depth d below the Earth's surface.
In simple words: As you go deeper into the Earth, the acceleration due to gravity decreases because only the mass closer to the center contributes to the gravitational pull, resulting in a linear decrease from the surface to the center.

🎯 Exam Tip: The crucial step in this derivation is realizing that only the mass of the sphere *inside* the radius (R-d) contributes to gravity at depth 'd'. The outer shell does not exert a net gravitational force. The final formula \(g_d = g(1 - d/R)\) shows a linear decrease with depth.

Question 12. State Kepler's three laws of planetary motion.
Answer:
- All planets move in elliptical orbits around the Sun with the Sun at one of the foci of the ellipse.
- The line that joins a planet and the Sun sweeps equal areas in equal intervals of time.
- The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semimajor axis of the ellipse traced by the planet.
In simple words: Kepler's laws describe how planets move: they orbit the Sun in ellipses, sweep equal areas in equal times (meaning they speed up closer to the Sun), and their orbital period squared is proportional to their average distance from the Sun cubed.

🎯 Exam Tip: Clearly state all three laws precisely. The first law defines the shape, the second describes the speed variation, and the third relates the period to the orbital size.

Question 13. State the formula for acceleration due to gravity at depth 'd' and altitude 'h' Hence show that their ratio is equal to \( \left(\frac{R-d}{R-2h}\right) \) by assuming that the altitude is very small as compared to the radius of the Earth.
Answer:
1. For an object at depth d, acceleration due to gravity of the Earth is given by,
\(g_d = g \left(1 - \frac{d}{R}\right)\) ....(1)
2. Also, the acceleration due to gravity at smaller altitude h is given by,
\(g_h = g \left(1 - \frac{2h}{R}\right)\) ....(2)
3. Hence, dividing equation (1) by equation (2),
we get,
\( \frac{g_d}{g_h} = \frac{g \left(1 - \frac{d}{R}\right)}{g \left(1 - \frac{2h}{R}\right)} \)
\( \therefore \frac{g_d}{g_h} = \frac{\frac{R-d}{R}}{\frac{R-2h}{R}} \)
\( \therefore \frac{g_d}{g_h} = \frac{R-d}{R-2h} \)
In simple words: The formula for gravity at depth 'd' is \(g_d = g(1 - d/R)\), and at a small altitude 'h' is \(g_h = g(1 - 2h/R)\). When we divide these, their ratio simplifies to \((R-d) / (R-2h)\).

🎯 Exam Tip: Ensure you use the correct approximate formula for gravity at small altitudes (\(g_h \approx g(1 - 2h/R)\)) and the exact formula for gravity at depth (\(g_d = g(1 - d/R)\)) before taking the ratio.

Question 14. What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
Answer: The exact horizontal velocity of projection that must be given to a satellite at a certain height so that it can revolve in a circular orbit round the Earth is called the critical velocity or orbital velocity (v_c).
Expression for critical velocity:
1. Consider a satellite of mass m revolving round the Earth at height h above its surface. Let M be the mass of the Earth and R be its radius.
2. If the satellite is moving in a circular orbit of radius (R + h) = r, its speed must be equal to the magnitude of critical velocity v_c.
3. The centripetal force necessary for circular motion of satellite is provided by gravitational force exerted by the satellite on the Earth.
\( \therefore \) Centripetal force = Gravitational force
\[ \frac{mv_c^2}{r} = \frac{GMm}{r^2} \]

\( \implies v_c^2 = \frac{GM}{r} \)

\( \implies v_c = \sqrt{\frac{GM}{r}} \)

\( \therefore v_c = \sqrt{\frac{GM}{(R+h)}} = \sqrt{g_h(R+h)} \)
This is the expression for critical speed at the orbit of radius (R + h).
4. The critical speed of a satellite is independent of the mass of the satellite. It depends upon the mass of the Earth and the height at which the satellite is revolving or gravitational acceleration at that altitude.
In simple words: Critical velocity is the precise horizontal speed a satellite needs to maintain a stable circular orbit around a planet, and it depends on the planet's mass and the orbit's radius.

🎯 Exam Tip: Remember to correctly equate centripetal force and gravitational force to derive the critical velocity expression, and clearly state its dependencies.

Question 15. Define escape speed. Derive an expression for the escape speed of an object from the surface of the earth.
Answer: 1. The minimum velocity with which a body should be thrown vertically upwards from the surface of the Earth so that it escapes the Earth's gravitational field, is called the escape velocity (v_e) of the body.
2. As the gravitational force due to Earth becomes zero at infinite distance, the object has to reach infinite distance in order to escape.
3. Let us consider the kinetic and potential energies of an object thrown vertically upwards with escape velocity v_e.
4. On the surface of the Earth,
\( K.E. = \frac{1}{2}mv_e^2 \)
\( P.E. = - \frac{GMm}{R} \)
Total energy = P.E. + K.E.
\( \therefore T.E. = \frac{1}{2}mv_e^2 - \frac{GMm}{R} \) .........(1)
5. The kinetic energy of the object will go on decreasing with time as it is pulled back by Earth's gravitational force. It will become zero when it reaches infinity. Thus, at infinite distance from the Earth,
K.E. = 0
Also,
\( P.E. = - \frac{GMm}{\infty} = 0 \)
\( \therefore \) Total energy = P.E. + K.E. = 0
6. As energy is conserved
\( \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \) ......[From(1)]

\( \implies \frac{1}{2}mv_e^2 = \frac{GMm}{R} \)

\( \implies v_e^2 = \frac{2GM}{R} \)
or, \( v_e = \sqrt{\frac{2GM}{R}} \)
In simple words: Escape speed is the minimum velocity an object needs to completely break free from Earth's gravity, calculated by equating its initial kinetic and potential energy to zero at an infinite distance.

🎯 Exam Tip: Focus on the conservation of mechanical energy principle and the concept of zero potential energy at infinity when deriving the escape velocity formula.

Question 16. Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth.
Answer: 1. Launching of a satellite in an orbit around the Earth cannot take place by use of single stage rocket. It requires minimum two stage rocket.
2. With the help of first stage of rocket, satellite can be taken to a desired height above the surface of the Earth.
3. Then the launcher is rotated in horizontal direction i.e., through 90° using remote control and the first stage of the rocket is detached.
4. With the help of second stage of rocket, a specific horizontal velocity (v_h) is given to satellite so that it can revolve in a circular path around the Earth.
5. The satellite follows different paths depending upon the horizontal velocity provided to it.
In simple words: A two-stage rocket is used to launch a satellite: the first stage lifts it to a certain height, then the second stage provides the necessary horizontal velocity to achieve orbit.

🎯 Exam Tip: When describing the launch process, highlight the role of each stage of the rocket and the importance of both altitude and horizontal velocity for achieving orbit.

4. Solve The Following Problems

Question 1. At what distance below the surface of the Earth, the acceleration due to gravity decreases by 10% of its value at the surface, given radius of Earth is 6400 km.
Answer: Solution:
Given: g_d = 90% of g i.e., \( \frac{g_d}{g} = 0.9 \),
R = 6400km = \( 6.4 \times 10^6 \) m
To find: Distance below the Earth's surface (d)
Formula: \( g_d = g(1 - \frac{d}{R}) \)
Calculation: From formula,
\( \frac{g_d}{g} = (1 - \frac{d}{R}) \)
\( \implies 0.9 = 1 - \frac{d}{R} \)
\( \implies \frac{d}{R} = 1 - 0.9 \)
\( \implies \frac{d}{R} = 0.1 \)
\( \implies d = R \times 0.1 \)
\( = 6.4 \times 10^6 \times 0.1 \)
\( = 640 \times 10^3 \) m
\( = 640 \) km
At distance 640 km below the surface of the Earth, value of acceleration due to gravity decreases by 10%.
In simple words: To find the depth where gravity decreases by 10%, we use the formula for gravitational acceleration inside the Earth and solve for the depth 'd' using the given Earth's radius.

🎯 Exam Tip: Remember the formula for acceleration due to gravity with depth. Pay attention to unit conversions (km to m) in numerical problems.

Question 2. If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
Answer: Solution:
Given: M_w = 10% of M \( = \frac{M}{10} \), D_w = D or R_w = R,
To find: Escape speed (v_ew)
Formula: \( v_e = \sqrt{\frac{2GM}{R}} \)
Calculation: From formula,
\( v_{ew} = \sqrt{\frac{2GM_w}{R_w}} \)
\( \therefore v_{ew} = \sqrt{\frac{2G(\frac{M}{10})}{R}} \)
\( \implies v_{ew} = \sqrt{\frac{1}{10} \times \frac{2GM}{R}} \)
\( \implies v_{ew} = \sqrt{\frac{1}{10}} \times v_e \)
\( \implies v_{ew} = v_e \times \frac{1}{\sqrt{10}} \)
As, we know that the escape speed from surface of the Earth is 11.2 km/s,
Substituting value of v_e = 11.2 km/s
\( v_{ew} = 11.2 \times \frac{1}{\sqrt{10}} = \frac{11.2}{3.162} \)
\( = 11.2 \times \frac{1}{3.162} \)
[Taking square root value]
= antilog {log(11.2) – Log(3.162)}
= antilog {1.0492 – 0.5000}
= antilog {0.5492} = 3.542
\( \therefore v_{ew} = 3.54 \)km/s
The escape velocity from the surface of wooden Earth is 3.54 km/s.
In simple words: If Earth's mass were 10% of its current value while its diameter remained the same, the escape speed would be reduced by a factor of \( \sqrt{10} \), resulting in a lower escape velocity.

🎯 Exam Tip: Remember the relationship between escape velocity, mass, and radius. For ratio problems, express the new quantity in terms of the original to simplify calculations.

Question 3. Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth.
Answer: Solution:
Given:- G = \( 6.67 \times 10^{-11} \) N m²/kg²
R = 6400 km
M = \( 6 \times 10^{24} \) kg
Given:- m = 2000 kg, h = 3600 km = \( 3.6 \times 10^6 \) m,
G = \( 6.67 \times 10^{-11} \) Nm²/kg²
R = 6400 km
M = \( 6 \times 10^{24} \) kg
To find: i) Kinetic energy (K.E.)
ii) Potential Energy (P.E.)
iii) Total Energy (T.E.)
iv) Binding Energy (B.E.)
Formulae: i. \( K.E. = \frac{GMm}{2(R+h)} \)
ii. \( P.E. = - \frac{GMm}{(R+h)} = -2(K.E.) \)
iii. T.E. = K.E. + P.E.
iv. B.E. = -T.E.
Calculation: From formula (i),
\[ K.E. = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 2 \times 10^3}{2 \times [(6.4 \times 10^6)+(3.6 \times 10^6)]} \]
\[ = \frac{6.67 \times 6 \times 10^{16}}{10^7} \]
\( = 40.02 \times 10^9 \) J
From formula (ii),
P.E. = \( -2 \times 40.02 \times 10^9 \)
\( = -80.04 \times 10^9 \) J
From formula (iii),
T.E. = \( (40.02 \times 10^9) + (-80.02 \times 10^9) \)
\( = -40.02 \times 10^9 \) J
From formula (iv),
B.E.= \( -(-40.02 \times 10^9) \)
\( = 40.02 \times 10^9 \) J
Kinetic energy of the satellite is \( 40.02 \times 10^9 \) J, potential energy is \( -80.04 \times 10^9 \) J, total energy is \( -40.02 \times 10^9 \) J and binding energy is \( 40.02 \times 10^9 \) J.
[Note: Total energy of orbiting satellite is negative.]
In simple words: For an orbiting satellite, kinetic energy is positive, potential energy is negative and twice the kinetic energy, total energy is the sum of these (negative), and binding energy is the positive equivalent of total energy.

🎯 Exam Tip: Remember the relationships between kinetic, potential, total, and binding energies for an orbiting satellite (e.g., K.E. = -T.E., P.E. = 2T.E., P.E. = -2K.E.).

Question 4. Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of satellite B is 4 \( \times 10^4 \) km. find radius of orbit of satellite A.
Answer: Solution:
Given: T_A = 1 hour, T_B = 8 hour,
r_B = \( 4 \times 10^4 \) km
To find: Radius of orbit of satellite A (r_A)
Formula: \( T = 2\pi \sqrt{\frac{r^3}{GM}} \)
Calculation: From formula,
\( T^2 = \frac{4\pi^2r^3}{GM} \)
\( \therefore T^2 \propto r^3 \)
\( \therefore (\frac{T_A}{T_B})^2 = (\frac{r_A}{r_B})^3 \)
\( \implies (\frac{1}{8})^2 = (\frac{r_A}{4 \times 10^4})^3 \)
\( \implies \frac{1}{64} = \frac{r_A^3}{(4 \times 10^4)^3} \)
\( \implies r_A^3 = \frac{1}{64} \times (4 \times 10^4)^3 \)
\( \implies r_A^3 = (\frac{4 \times 10^4}{4})^3 \)
\( \implies r_A^3 = (1 \times 10^4)^3 \)
\( \implies r_A = 1 \times 10^4 \) km
Radius of orbit of satellite A will be \( 1 \times 10^4 \) km.
In simple words: Using Kepler's Third Law, which states that the square of the orbital period is proportional to the cube of the orbital radius, we can find the unknown orbital radius of satellite A given the periods and one radius.

🎯 Exam Tip: Remember Kepler's Third Law (T² \( \propto \) r³) for problems involving orbital periods and radii. Ensure consistent units for time and distance.

Question 5. Find the gravitational force between the Sun and the Earth.
Answer: Solution:
Given Mass of the Sun = \( 1.99 \times 10^{30} \) kg
Mass of the Earth = \( 5.98 \times 10^{24} \) kg
The average distance between the Earth and the Sun = \( 1.5 \times 10^{11} \) m.
Solution:
Given: M_S = \( 1.99 \times 10^{30} \) kg
M_E = \( 5.98 \times 10^{24} \) kg, R = \( 1.5 \times 10^{11} \) m.
To find: Gravitational force between the Sun and the Earth (F)
Formula: \( F = \frac{Gm_1 m_2}{r^2} \)
Calculation: As, we know, G = \( 6.67 \times 10^{-11} \) N m²/kg²
From formula,
\( F = \frac{GM_E M_S}{R^2} \)
\[ F = \frac{6.67 \times 10^{-11} \times 1.99 \times 10^{30} \times 5.98 \times 10^{24}}{(1.5 \times 10^{11})^2} \]
\[ = \frac{6.67 \times 1.99 \times 5.98}{2.25} \times 10^{21} \]
= antilog {(log(6.67) + log( 1.99) + log(5.98) – log(2.25))} \( \times 10^{21} \)
= antilog {(0.8241) + (0.2989) + (0.7767) – (0.3522)} \( \times 10^{21} \)
= antilog {1.5475} \( \times 10^{21} \)
= \( 35.28 \times 10^{21} \)
\( = 3.5 \times 10^{22} \) N
The gravitational force between the Sun and the Earth is \( 3.5 \times 10^{22} \) N.
In simple words: The gravitational force between the Sun and Earth is calculated using Newton's Law of Universal Gravitation, plugging in their masses, the distance between them, and the gravitational constant.

🎯 Exam Tip: Ensure correct values for masses, distance, and the universal gravitational constant. Double-check scientific notation and exponent calculations.

Question 6. Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M = 5.98 \( \times 10^{24} \) kg, R = 6400 km).
Answer: Solution:
Given: h = 300 km = \( 0.3 \times 10^6 \) m,
M = \( 5.98 \times 10^{24} \) kg,
R = 6400km = \( 6.4 \times 10^6 \) m
G = \( 6.67 \times 10^{-11} \) N m²/kg²
To find: Acceleration due to gravity at height (g_h)
Formula: \( g_h = \frac{GM}{(R+h)^2} \)
Calculation: From formula,
\[ g_h = \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{[(6.4 \times 10^6)+(0.3 \times 10^6)]^2} \]
\[ = \frac{6.67 \times 5.98 \times 10^{13}}{(6.7)^2 \times 10^{12}} \]
\( = 6.67 \times 10^{11} \times 5.98 \times 10^{11} \)
= antilog {log(6.67) + log(5.98) – 2log(6.7)} \( \times 10 \)
= antilog{0.8241 + 0.7767 – 2(0.8261)} \( \times 10 \)
= antilog {1.6008 – 1.6522} \( \times 10 \)
= antilog {\(\bar{1}.9486 \)} \( \times 10 \)
\( = 0.8884 \times 10 = 8.884 \) m/s²
Acceleration due to gravity at 300 km will be 8.884 m/s².
In simple words: We calculate the acceleration due to gravity at a specific height above the Earth's surface by using the formula that accounts for the increased distance from the Earth's center.

🎯 Exam Tip: Ensure precise calculations for (R+h)² and proper handling of exponents. Unit consistency is crucial (convert km to m for height and radius). The formula \( g_h = \frac{GM}{(R+h)^2} \) is key.

Question 7. Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth's surface. M_E = 5.98 \( \times 10^{24} \) kg, R = 6.4 \( \times 10^6 \) m.
Answer: Solution:
Given: h = 1000 km = \( 1 \times 10^6 \) m,
M_E = \( 5.98 \times 10^{24} \) kg, R = \( 6.4 \times 10^6 \) m,
G = \( 6.67 \times 10^{-11} \) N m²/kg²
To find: Speed of satellite (v_c)
Formula: \( v_c = \sqrt{\frac{GM}{r}} \)
Calculation:
From formula,
\[ v_c = \sqrt{\frac{GM}{(R+h)}} \]
\[ = \sqrt{\frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{[(6.4 \times 10^6) + (1 \times 10^6)]}} \]
\[ = \sqrt{\frac{6.67 \times 5.98 \times 10^{13}}{7.4 \times 10^6}} \]
\[ = \sqrt{\frac{6.67 \times 5.98}{7.4} \times 10^7} \]
= antilog{log(6.67) + log(5.98) - log(7.4) \( \times 10^7 \)}
= antilog{0.8241 + 0.7767 – 0.8692} \( \times 10^7 \)
= antilog{0.7316} \( \times 10^7 \)
\( = \sqrt{5.391 \times 10^7} \)
\( = \sqrt{53.91 \times 10^6} \)
\( = 7.343 \times 10^3 \) .... [Taking square root value]
\( = 7.34 \times 10^3 \) m/s
Speed of the satellite at height 1000 km is \( 7.34 \times 10^3 \) m/s.
In simple words: The speed of a satellite in orbit depends on the gravitational constant, the mass of the central body (Earth), and the radius of its orbit (Earth's radius plus orbital height).

🎯 Exam Tip: Remember to use the total orbital radius (R+h) in the critical velocity formula. Be careful with unit conversions and exponent handling during calculation.

Question 8. Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars = 3.4 \( \times 10^3 \) km and its mass is 6.4 \( \times 10^{23} \) kg.
Answer: Solution:
Given:
M = \( 6.4 \times 10^{23} \) kg
R = \( 3.4 \times 10^3 \) km = \( 3.4 \times 10^6 \) m,
To find: Acceleration due to gravity on the surface of the Mars (g_M)
Formula: \( g = \frac{GM}{R^2} \)
Calculation: As, G = \( 6.67 \times 10^{-11} \) N m²/kg²
From formula,
\[ g_M = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{(3.4 \times 10^6)^2} \]
\[ = \frac{6.67 \times 6.4}{3.4 \times 3.4} \]
= antilog {log(6.67) + log(6.4) – log(3.4) – log(3.4)}
= antilog {(0.8241) + (0.8062) – (0.5315) – (0.53 15)}
= antilog {0.5673}
\( = 3.693 \) m/s²
Acceleration due to gravity on the surface of Mars is 3.693 m/s².
In simple words: The acceleration due to gravity on Mars is calculated using Newton's law of gravitation, applying Mars's mass and radius along with the universal gravitational constant.

🎯 Exam Tip: Ensure proper unit conversions for radius (km to m) and use the correct formula \( g = \frac{GM}{R^2} \) for acceleration due to gravity on a planet's surface.

Question 9. A planet has mass 6.4 \( \times 10^{24} \) kg and radius 3.4 \( \times 10^6 \) m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity.
Answer: Solution:
Given: M = \( 6.4 \times 10^{24} \) kg, R = \( 3.4 \times 10^6 \) m, m = 800 kg
To find: Energy required to remove the object from surface of planet to infinity = B.E.
Formula: \( B.E. = \frac{GMm}{R} \)
Calculation: We know that,
G = \( 6.67 \times 10^{-11} \) N m²/kg²
From formula,
\[ B.E. = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{24} \times 800}{3.4 \times 10^6} \]
\[ B.E. = \frac{6.67 \times 6.4 \times 8}{3.4} \times 10^9 \]
\[ = \frac{6.67 \times 51.2}{3.4} \times 10^9 \]
= antilog{log(6.67) + log(51.2) – log(3.4)} \( \times 10^9 \)
= antilog{0.8241 + 1.7093 – 0.5315} \( \times 10^9 \)
= antilog {2.0019} \( \times 10^9 \)
\( = 1.004 \times 10^2 \times 10^9 \)
\( = 1.004 \times 10^{11} \) J
Energy required to remove the object from the surface of the planet is \( 1.004 \times 10^{11} \) J.
[Note: Answer calculated above is in accordance with textual methods of calculation.]
In simple words: The energy required to remove an object from a planet's surface to infinity is its binding energy, calculated using the gravitational potential energy formula with the planet's mass and radius, and the object's mass.

🎯 Exam Tip: For binding energy calculations, use the formula \( B.E. = \frac{GMm}{R} \). Ensure all given values (masses, radius, G) are in SI units before calculation.

Question 10. Calculate the value of the universal gravitational constant from the given data. Mass of the Earth = 6 \( \times 10^{24} \) kg, Radius of the Earth = 6400 km and the acceleration due to gravity on the surface = 9.8 m/s²
Answer: Solution:
Given: M = \( 6 \times 10^{24} \) kg,
R = 6400km = \( 6.4 \times 10^6 \) m,
g = 9.8 m/s²
To find: Gravitational constant (G)
Formula. \( g = \frac{GM}{R^2} \)
Calculation: From formula,
\( G = \frac{gR^2}{M} \)
\[ G = \frac{9.8 \times (6.4 \times 10^6)^2}{6 \times 10^{24}} \]
\[ = \frac{9.8 \times 40.96 \times 10^{12}}{6 \times 10^{24}} \]
\[ = \frac{401.408 \times 10^{12}}{6 \times 10^{24}} \]
\[ = \frac{401.4 \times 10^{12}}{6 \times 10^{24}} \]
\( \therefore G = 6.69 \times 10^{-11} \) N m²/kg²
The value of gravitational constant is \( 6.69 \times 10^{-11} \) N m²/kg².
In simple words: The universal gravitational constant can be determined by rearranging the formula for acceleration due to gravity, using the Earth's mass, radius, and surface gravity.

🎯 Exam Tip: Recall the formula \( g = \frac{GM}{R^2} \) and rearrange it to solve for G. Ensure all units are consistent (SI) for accurate calculation.

Question 11. A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 1/7 times the mass of the Earth and radius twice that of the Earth's radius.
Answer: Solution:
Given: W_E = 5.6 kg-wt.,
\( \frac{M_P}{M_E} = \frac{1}{7} \), \( \frac{R_P}{R_E} = 2 \)
To find: Weight of the body on the surface of planet (W_P)
Formula: W = mg \( = \frac{GMm}{R^2} \)
Calculation: From formula,
\[ \frac{W_P}{W_E} = \frac{\frac{GM_P m}{R_P^2}}{\frac{GM_E m}{R_E^2}} \]
\[ \frac{W_P}{W_E} = \frac{M_P}{M_E} \times (\frac{R_E}{R_P})^2 \]
\[ \frac{W_P}{5.6} = \frac{1}{7} \times (\frac{1}{2})^2 \]
\[ \frac{W_P}{5.6} = \frac{1}{7} \times \frac{1}{4} \]
\[ \frac{W_P}{5.6} = \frac{1}{28} \]
\( \therefore W_P = \frac{1}{28} \times 5.6 = 0.2 \) kg-wt.
Weight of the body on the surface of a planet will be 0.2 kg-wt.
[Note: The answer given above is calculated in accordance with textual method considering the given data].
In simple words: The weight of a body on a different planet can be calculated by comparing the ratios of its mass and radius to Earth's, which affects the acceleration due to gravity.

🎯 Exam Tip: Remember that weight depends on the acceleration due to gravity (g), which itself depends on the planet's mass and radius. Use ratio comparisons to simplify calculations.

Question 12. What is the gravitational potential due to the Earth at a point which is at a height of 2R_E above the surface of the Earth, Mass of the Earth is 6 \( \times 10^{24} \) kg, radius of the Earth = 6400 km and G = 6.67 \( \times 10^{-11} \) Nm² kg-².
Answer: Solution:
Given: M = \( 6 \times 10^{24} \) kg,
R_E = 6400km = \( 6.4 \times 10^6 \) m,
G = \( 6.67 \times 10^{-11} \) Nm²/kg²,
h = 2R_E
To find: Gravitational potential (V)
Formula: \( V = - \frac{GM}{r} \)
Calculation: From formula,
\[ V = - \frac{GM}{(R_E+2R_E)} \]
\[ = - \frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{3 \times 6.4 \times 10^6} \]
\[ = - \frac{6.67 \times 2}{6.4} \times 10^7 \]
\( = -2.08 \times 10^7 \) J kg\(^{-1}\)
Negative sign indicates the attractive nature of gravitational potential.
Gravitational potential due to Earth will be \( 2.08 \times 10^7 \) J kg\(^{-1}\) towards the centre of the Earth.
[Note: According to definition of gravitational potential its SI unit is J/kg.]
In simple words: Gravitational potential at a point is the work done per unit mass to bring a test mass from infinity to that point, and it is calculated using the gravitational constant, Earth's mass, and the distance from the Earth's center.

🎯 Exam Tip: Remember that gravitational potential is a scalar quantity and is always negative. The total distance 'r' is the sum of Earth's radius and the given height from the surface.

Can You Recall? (Textbook Page No. 78)

Question 1. i) What are Kepler's laws?
ii)What is the shape of the orbits of planets?
Answer: 1. The Kepler's laws are:
(i) Kepler's first law: The orbit of a planet is an ellipse with the Sun at one of the foci.
(ii) Kepler's second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
(iii) Kepler's third law: The square of orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun.
2. The orbits of the planet are elliptical in shape.
In simple words: Kepler's laws describe planetary motion: planets orbit the Sun in ellipses (first law), sweep equal areas in equal times (second law), and the square of their orbital period is proportional to the cube of their average distance from the Sun (third law), meaning planetary orbits are elliptical.

🎯 Exam Tip: Clearly state all three of Kepler's laws. For the first law, mention the elliptical shape and the Sun's position; for the second, equal areas; and for the third, the mathematical proportionality (T² \( \propto \) r³).

Question 2. When released from certain height why do objects tend to fall vertically downwards?
Answer: When released from certain height, objects tend to fall vertically downwards because of the gravitational force exerted by the Earth.
In simple words: Objects fall vertically downwards because Earth's gravitational force attracts them directly towards its center.

🎯 Exam Tip: Keep the explanation simple and direct, focusing on Earth's gravitational pull as the primary reason.