Question 1

Q1. You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside

Accepted Answer

No, a body cannot be shielded from the gravitational influence of nearby matter by putting it inside a hollow sphere or by any other known means.

Here's why:
1.  **Nature of Forces:** Electrical forces can be attractive or repulsive, depending on the sign of the charges. This allows for the possibility of 'screening' or 'shielding' by arranging charges (e.g., in a hollow conductor, free electrons redistribute to cancel the internal electric field). Gravitational force, however, is always attractive and depends only on mass. There is no 'negative mass' or 'gravitational charge' that could be used to cancel out a gravitational field.
2.  **Field Properties:** The electric field inside a hollow conductor due to external charges is zero (electrostatic shielding). This is a consequence of Gauss's law and the ability of charges to move freely within the conductor. For gravity, there is no analogous mechanism. The gravitational field inside a hollow sphere due to masses *outside* is generally not zero. Only if the hollow sphere itself is the source of gravity (and is uniform), then the gravitational field *inside* due to the sphere's own mass is zero. But this doesn't shield from *external* gravitational influences.
3.  **Lack of Dipoles:** Gravitational fields do not exhibit dipole or multipole shielding effects like electric fields. Gravitational fields are solely due to mass, and mass is always positive.

Question 2

Q2. An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the spaceship is too small to create a detectable gravitational field, what is the reason

Accepted Answer

The astronaut feels weightless not because there is no gravity, but because both the astronaut and the spaceship are in a continuous state of free fall around the Earth.

Here's a detailed explanation:
1.  **Presence of Gravity:** Even in orbit, at an altitude of a few hundred kilometers, the Earth's gravitational field is still quite strong (e.g., at 400 km altitude, g is about 90% of its value on the surface). So, gravity is definitely present and acting on both the astronaut and the spaceship.
2.  **Free Fall:** The spaceship is constantly falling towards the Earth, but its high tangential velocity prevents it from hitting the Earth, causing it to orbit. The astronaut inside the spaceship is also falling towards the Earth with the *same acceleration* as the spaceship.
3.  **Absence of Normal Force:** Since both the astronaut and the spaceship are accelerating together (falling together) towards the Earth, there is no relative acceleration between them. The astronaut does not press against the floor of the spaceship, and consequently, the spaceship does not exert a normal force on the astronaut. It is this absence of a normal force (which we perceive as our 'weight' when standing on a surface) that creates the sensation of weightlessness.
4.  **Apparent Weight:** Weightlessness is therefore an 'apparent' phenomenon, a consequence of being in a continuously falling frame of reference, rather than an actual absence of gravitational force.

Question 3

Q3. The gravitational force between two objects is proportional to 1/R (where R is the separation between them) instead of 1/R². Then, the orbital velocity of a satellite in a circular

Accepted Answer

The correct option is (A) R⁰.

Question 4

Q4. A body is weighed by a spring balance in a spaceship sailing at a constant velocity of 1000 km/h towards the sun. Would the reading of the spring balance change from that on the ea

Accepted Answer

Yes, the reading of the spring balance would change from that on the Earth.

Here's the explanation:
1.  **What a Spring Balance Measures:** A spring balance measures the apparent weight of an object, which is essentially the force required to support the object against gravity. This force is directly related to the local gravitational field strength.
2.  **Effect of Constant Velocity:** The fact that the spaceship is moving at a *constant velocity* (1000 km/h) is crucial. According to Newton's laws, an object moving at a constant velocity is in an inertial frame of reference. In such a frame, there are no fictitious forces (like those experienced in an accelerating frame) that would alter the apparent weight due to the motion itself. So, the constant velocity *itself* does not cause a change in the spring balance reading.
3.  **Change in Gravitational Environment:** The key factor here is the change in the *gravitational environment*. When the spaceship is 'sailing towards the sun', it implies it is moving away from the Earth. As the spaceship moves away from Earth, the gravitational force exerted by the Earth on the body decreases significantly (as F ∝ 1/R²). While the gravitational force from the Sun would be present, it is much weaker than Earth's surface gravity (g ≈ 9.8 m/s²) and also much weaker than Earth's gravity at orbital altitudes.
4.  **Conclusion:** Because the net gravitational force acting on the body in the spaceship (far from Earth) would be much less than the gravitational force on the body at the Earth's surface, the spring balance would show a significantly lower reading compared to its reading on Earth. If the spaceship were in deep space, far from any significant gravitational source, the reading would approach zero.

Question 5

Q5. A man can jump higher on the moon than on the earth. Why?

Accepted Answer

A man can jump higher on the Moon than on the Earth primarily because the acceleration due to gravity on the Moon is significantly less than that on Earth.

Here's a detailed explanation:
1.  **Acceleration due to Gravity (g):** The acceleration due to gravity on the Moon (g_moon) is approximately one-sixth of the acceleration due to gravity on Earth (g_earth). That is, g_moon ≈ g_earth / 6.
2.  **Initial Jump Velocity:** When a person jumps, the initial upward velocity (let's call it 'u') they achieve is determined by their muscular strength and effort. This initial velocity 'u' would be roughly the same whether they are on Earth or on the Moon, as their muscular strength doesn't change.
3.  **Kinematic Equation for Height:** For a projectile launched vertically upwards with an initial velocity 'u' under constant acceleration 'g', the maximum height 'h' reached can be found using the kinematic equation: v² = u² + 2gh. At the maximum height, the final velocity (v) is 0. Taking upward as positive, 'g' acts downwards, so it's negative:
    0 = u² - 2gh
    Rearranging for h:
    h = u² / (2g)
4.  **Comparison:** Since the initial velocity 'u' is approximately the same on both Earth and Moon, and 'g_moon' is much smaller than 'g_earth', the maximum height 'h' achieved on the Moon will be proportionally greater.
    h_moon = u² / (2 * g_moon)
    h_earth = u² / (2 * g_earth)
    Since g_moon ≈ g_earth / 6, we have:
    h_moon = u² / (2 * (g_earth / 6)) = 6 * (u² / (2 * g_earth)) = 6 * h_earth
    Therefore, a man can jump approximately six times higher on the Moon than on the Earth.

Question 6

Q6. A rocket is fired from the Earth towards the Sun. At what distance from the Earth’s centre is the gravitational force on the rocket zero? Mass of the Sun = 2 × 10^30 kg, Mass of th

Accepted Answer

The gravitational force on the rocket will be zero at a point where the gravitational pull of the Earth is exactly balanced by the gravitational pull of the Sun. Let this point be at a distance 'x' from the Earth's centre. The distance of this point from the Sun's centre will then be (R - x), where R is the orbital radius of the Earth around the Sun. By equating the gravitational forces, we can find 'x'.

The distance from the Earth's centre where the gravitational force on the rocket is zero is approximately 2.59 × 10^8 m.

Question 7

Q7. How will you ‘weigh’ the Sun, that is, estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 × 10^8 km.

Accepted Answer

We can estimate the mass of the Sun by considering the orbital motion of the Earth around it. The gravitational force exerted by the Sun on the Earth provides the necessary centripetal force for the Earth's orbital motion. By equating these two forces and using the known orbital parameters of the Earth, we can calculate the Sun's mass.

The estimated mass of the Sun is approximately 2.0 × 10^30 kg.

Question 8

Q8. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 × 10^8 km away from the Sun?

Accepted Answer

This problem can be solved using Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of its semi-major axis (R) (or mean orbital radius) around the Sun (T^2 ∝ R^3). This means that the ratio T^2/R^3 is constant for all planets orbiting the Sun.

Using Kepler's Third Law, the distance of Saturn from the Sun is approximately 1.43 × 10^12 m or 1.43 × 10^9 km.

Question 9

Q9. A body weighs 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?

Accepted Answer

The weight of a body is the gravitational force acting on it. The acceleration due to gravity (g) changes with height above the Earth's surface. We can use the formula for 'g' at a certain height to find the new gravitational force.

The gravitational force on the body at a height equal to half the radius of the Earth is 28 N.

Question 10

Q10. Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the Earth if it weighed 250 N on the surface?

Accepted Answer

The weight of a body depends on the acceleration due to gravity (g). When a body is taken inside the Earth, the value of 'g' changes with depth. For a uniform mass density Earth, the acceleration due to gravity at a depth 'd' from the surface (or distance r from the centre) is directly proportional to the distance from the centre.

The body would weigh 125 N half way down to the centre of the Earth.

Question 11

Q11. A rocket is fired from the Earth towards the Sun. At what distance from the Earth's centre is the gravitational force on the rocket zero? Mass of the Sun = 2 × 10^30 kg, Mass of th

Accepted Answer

The gravitational force on the rocket will be zero at a point where the gravitational force exerted by the Earth is equal in magnitude and opposite in direction to the gravitational force exerted by the Sun. This point lies between the Earth and the Sun. The distance from the Earth's centre where the net gravitational force is zero is approximately 2.59 × 10^8 m.

Question 12

Q12. How will you ‘weigh’ the Sun, that is, estimate its mass? The mean orbital radius of the Earth around the Sun is 1.5 × 10^8 km.

Accepted Answer

To estimate the mass of the Sun, we can use the Earth's orbital motion around the Sun. Assuming a circular orbit, the gravitational force exerted by the Sun on the Earth provides the necessary centripetal force for the Earth's orbit. By equating these two forces and using the known orbital radius and period of the Earth, we can calculate the Sun's mass. The estimated mass of the Sun is approximately 2.0 × 10^30 kg.

Question 13

Q13. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 × 10^8 km away from the Sun?

Accepted Answer

This problem can be solved using Kepler's Third Law of Planetary Motion, which states that the square of the orbital period (T) of a planet is directly proportional to the cube of its semi-major axis (R) (or mean orbital radius for circular orbits). The distance of Saturn from the Sun is approximately 1.43 × 10^9 km.

Question 14

Q14. A body weighs 63 N on the surface of the Earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of the Earth?

Accepted Answer

The weight of a body is the gravitational force acting on it. The acceleration due to gravity decreases with height above the Earth's surface. At a height equal to half the radius of the Earth, the gravitational force on the body will be approximately 28 N.

Question 15

Q15. Assuming the Earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the Earth if it weighed 250 N on the surface?

Accepted Answer

The weight of a body is the gravitational force acting on it. The acceleration due to gravity decreases linearly with depth inside the Earth, assuming uniform mass density. Halfway down to the centre of the Earth, the body would weigh 125 N.

Question 16

Q16. A satellite orbits the Earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of Earth's gravitational influence? Mass of the s

Accepted Answer

To rocket the satellite out of Earth's gravitational influence, energy equal to the negative of its total mechanical energy in orbit must be supplied. The total mechanical energy of a satellite in a circular orbit is given by E = -GMm / (2r), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the orbital radius.

Given:
Height of satellite, h = 400 km = 400 × 10^3 m
Mass of satellite, m = 200 kg
Mass of Earth, M = 6.0 × 10^24 kg
Radius of Earth, R = 6.4 × 10^6 m
Gravitational constant, G = 6.67 × 10^-11 N m^2 kg^-2

First, calculate the orbital radius (r):
r = R + h = 6.4 × 10^6 m + 400 × 10^3 m = 6.4 × 10^6 m + 0.4 × 10^6 m = 6.8 × 10^6 m

Next, calculate the total mechanical energy (E) of the satellite in orbit:
E = -GMm / (2r)
E = -(6.67 × 10^-11 N m^2 kg^-2) × (6.0 × 10^24 kg) × (200 kg) / (2 × 6.8 × 10^6 m)
E = -(6.67 × 6.0 × 200 × 10^(-11+24)) / (13.6 × 10^6) J
E = -(8004 × 10^13) / (13.6 × 10^6) J
E = -(8004 / 13.6) × 10^(13-6) J
E = -588.53 × 10^7 J
E ≈ -5.89 × 10^9 J

The energy required to rocket the satellite out of Earth's gravitational influence is the negative of this total energy:
Energy required = -E = -(-5.89 × 10^9 J) = 5.89 × 10^9 J

Therefore, approximately 5.89 × 10^9 J of energy must be expended.

Question 17

Q17. Two stars each of one solar mass ( = 2 × 10^30 kg) are approaching each other for a head on collision. When they are a distance 10^9 km apart, their speeds are negligible. What is

Accepted Answer

This problem can be solved using the principle of conservation of mechanical energy. The system consists of two stars, and only gravitational forces (conservative forces) are acting. Therefore, the total mechanical energy (Kinetic Energy + Potential Energy) of the system remains constant.

Given:
Mass of each star, M1 = M2 = M = 1 solar mass = 2 × 10^30 kg
Initial distance between centers, r_initial = 10^9 km = 10^9 × 10^3 m = 10^12 m
Initial speeds, v_initial = 0 (negligible)
Radius of each star, R_star = 10^4 km = 10^4 × 10^3 m = 10^7 m
Gravitational constant, G = 6.67 × 10^-11 N m^2 kg^-2

**1. Initial State (Stars at 10^12 m apart):**
Initial Kinetic Energy, KE_initial = 0 (since speeds are negligible)
Initial Gravitational Potential Energy, PE_initial = -G M1 M2 / r_initial
PE_initial = -(6.67 × 10^-11) × (2 × 10^30) × (2 × 10^30) / (10^12)
PE_initial = -(6.67 × 4 × 10^(-11 + 30 + 30 - 12))
PE_initial = -26.68 × 10^37 J

Total Initial Energy, E_initial = KE_initial + PE_initial = 0 + (-26.68 × 10^37 J) = -26.68 × 10^37 J

**2. Final State (Stars Collide):**
When the stars collide, their surfaces touch. The distance between their centers at collision (r_final) will be the sum of their radii.
r_final = R_star + R_star = 2 × R_star = 2 × 10^7 m

Let the speed of each star just before collision be v. Since the stars are identical and approaching each other symmetrically, they will have equal speeds relative to their center of mass. The total kinetic energy of the system will be the sum of the kinetic energies of the two stars.
Final Kinetic Energy, KE_final = (1/2) M v^2 + (1/2) M v^2 = M v^2
KE_final = (2 × 10^30) v^2

Final Gravitational Potential Energy, PE_final = -G M1 M2 / r_final
PE_final = -(6.67 × 10^-11) × (2 × 10^30) × (2 × 10^30) / (2 × 10^7)
PE_final = -(6.67 × 4 × 10^(-11 + 30 + 30)) / (2 × 10^7)
PE_final = -(26.68 × 10^49) / (2 × 10^7)
PE_final = -13.34 × 10^(49 - 7) J
PE_final = -13.34 × 10^42 J

Total Final Energy, E_final = KE_final + PE_final = (2 × 10^30) v^2 - (13.34 × 10^42) J

**3. Apply Conservation of Energy:**
E_initial = E_final
-26.68 × 10^37 = (2 × 10^30) v^2 - (13.34 × 10^42)

Rearrange to solve for v^2:
(2 × 10^30) v^2 = 13.34 × 10^42 - 26.68 × 10^37
(2 × 10^30) v^2 = 13.34 × 10^42 - 0.0002668 × 10^42
(2 × 10^30) v^2 ≈ 13.34 × 10^42 (since 26.68 × 10^37 is very small compared to 13.34 × 10^42)

v^2 = (13.34 × 10^42) / (2 × 10^30)
v^2 = 6.67 × 10^(42 - 30)
v^2 = 6.67 × 10^12

v = √(6.67 × 10^12)
v = √(66.7 × 10^11)  (or √(6.67) × 10^6)
v ≈ 2.583 × 10^6 m/s

The speed with which each star collides is approximately 2.58 × 10^6 m/s.

Question 18

Q18. The gravitational intensity at the centre of a hemispherical shell of uniform mass density is directed along the axis of symmetry. Using this fact, show that the force on a point m

Accepted Answer

We can derive the force on the point mass m by integrating the gravitational force exerted by infinitesimal elements of the hemispherical shell. The result shows the force is F = GMm / (2R²).

**Derivation:**
1.  **Setup:** Consider a thin hemispherical shell of mass M and radius R. A point mass 'm' is placed at its center O. Let the axis of symmetry be the y-axis.
2.  **Mass Density:** The surface area of the hemisphere is 2πR². The surface mass density is σ = M / (2πR²).
3.  **Elemental Ring:** Consider a small elemental ring on the surface of the hemisphere at an angle θ with the y-axis and of angular thickness dθ. The radius of this ring is r = Rsinθ and its thickness is Rdθ.
4.  **Mass of the Ring (dM):** The area of this ring is dA = (circumference) × (thickness) = (2πr)(Rdθ) = 2π(Rsinθ)Rdθ = 2πR²sinθdθ. The mass of this ring is dM = σ dA = (M / 2πR²) × (2πR²sinθdθ) = Msinθdθ.
5.  **Force by the Ring (dF):** Every point on this ring is at a distance R from the center O. The gravitational force exerted by this ring on the mass 'm' at the center is dF = G(dM)m / R².
6.  **Symmetry and Components:** Due to symmetry, the components of the force perpendicular to the axis of symmetry (y-axis) will cancel out. Only the components along the axis of symmetry will add up. The component of dF along the y-axis is dF_y = dF cosθ.
7.  **Integration:** The total force F is the integral of dF_y over the entire hemisphere (from θ = 0 to θ = π/2).
    F = ∫ dF_y = ∫ (G(dM)m / R²) cosθ
    F = ∫ (from 0 to π/2) [G(Msinθdθ)m / R²] cosθ
    F = (GMm / R²) ∫ (from 0 to π/2) sinθcosθ dθ
8.  **Solving the Integral:** The integral can be solved as follows:
    ∫ sinθcosθ dθ = (1/2) ∫ sin(2θ) dθ = (1/2) [-cos(2θ)/2] = -cos(2θ)/4.
    Evaluating from 0 to π/2:
    [-cos(π)/4] - [-cos(0)/4] = [-(-1)/4] - [-1/4] = 1/4 + 1/4 = 1/2.
    Alternatively, using substitution u = sinθ, du = cosθdθ. The limits change from 0 to 1. ∫ (from 0 to 1) u du = [u²/2] (from 0 to 1) = 1/2.
9.  **Final Result:**
    F = (GMm / R²) × (1/2) = GMm / (2R²).
    This force is directed along the axis of symmetry, away from the shell's base.

**Case of a Full Sphere:**
If the shell were a full sphere, the force on the point mass 'm' placed at the center would be **zero**. This is a direct consequence of the shell theorem, which states that the gravitational field inside a uniform spherical shell is zero everywhere. The forces from all parts of the shell cancel each other out perfectly at the center.

Question 19

Q19. A rocket is fired from the Earth towards the Sun. At what distance from the Earth's centre is the gravitational force on the rocket zero? Mass of the Sun = 2 × 10^30 kg, Mass of th

Accepted Answer

The gravitational force on the rocket will be zero at a distance of approximately 2.59 × 10^8 m from the Earth's centre.

Question 20

Q20. How will you ‘weigh’ the Sun, that is, estimate its mass? The average orbital radius of the Earth around the Sun is 1.5 × 10^11 m.

Accepted Answer

To estimate the mass of the Sun, we can use the principles of gravitational force and centripetal force, considering the Earth's orbital motion around the Sun. The gravitational force exerted by the Sun on the Earth provides the necessary centripetal force for the Earth to maintain its orbit. By equating these two forces and using the known orbital radius and period of the Earth, we can derive an expression for the Sun's mass.

Question 21

Q21. A Saturn year is 29.5 times the Earth year. How far is the Saturn from the Sun if the Earth is 1.50 × 10^8 km away from the Sun?

Accepted Answer

Saturn is approximately 1.43 × 10^9 km away from the Sun.

Chapter 8: Gravitation