CBSE Class 10 Physics Gravitation MCQs Set 23

Answer: (c) between the centres of the bodies

Explanation:
1. For spherical bodies the force acts along the line joining their centres.
2. Distance r is the centre-to-centre separation.

Teacher's Note:
1. For a satellite, r = R + h from the centre of the earth.

Answer: (d) 7.9 km/s

Explanation:
1. \( v = \sqrt{gR} = \sqrt{9.8 \times 6.4\times10^6} \approx 7920 \) m/s.
2. = about 7.9 km/s.

Teacher's Note:
1. 11.2 km/s is the escape speed, not the orbital speed.

Answer: (b) greater at the poles than at the equator

Explanation:
1. The earth is slightly flattened, so the radius is smaller at the poles.
2. Also the rotation of the earth reduces g at the equator.
3. Hence g is greatest at the poles.

Teacher's Note:
1. g increases as we move from the equator to the poles.

Answer: (a) one-fourth

Explanation:
1. At h = R the distance from the centre is 2R.
2. Force becomes 1/4 of the surface weight.

Teacher's Note:
1. Inverse-square dependence on the distance from the earth's centre.

Answer: (c) is the same for all bodies if air resistance is neglected

Explanation:
1. \( g = GM/R^2 \) does not contain the mass of the falling body.
2. In vacuum a feather and a stone fall together.

Teacher's Note:
1. Air resistance, not gravity, makes light objects fall slower in air.

Answer: (c) the gravitational pull provides the centripetal force needed for its circular motion

Explanation:
1. The moon has a tangential velocity.
2. The earth's gravity supplies the centripetal force that keeps changing the direction of this velocity, so the moon keeps falling around the earth.

Teacher's Note:
1. Without the tangential velocity the moon would fall straight down.
2. Without gravity it would fly off along a tangent.

Answer: (a) 2 s

Explanation:
1. \( h = \tfrac12 g t^2 \): \( 19.6 = 4.9\,t^2 \).
2. \( t^2 = 4 \), so t = 2 s.

Teacher's Note:
1. Free fall from rest: \( t = \sqrt{2h/g} \).

Answer: (b) 70 kg

Explanation:
1. Mass does not depend on g.
2. It stays 70 kg on the moon.

Teacher's Note:
1. Weight on the moon would be about 114 N, not 70 kg.

Answer: (a) 4g

Explanation:
1. \( g \propto M/R^2 \).
2. With R halved, g becomes \( 1/(1/2)^2 = 4 \) times, i.e. 4g.

Teacher's Note:
1. Halving the radius quadruples g for the same mass.

Answer: (a) decreases as the orbit gets larger

Explanation:
1. \( v = \sqrt{GM/(R+h)} \).
2. A larger orbital radius gives a smaller speed.

Teacher's Note:
1. Satellites in higher orbits move more slowly and have longer periods.

Answer: (c) remains unchanged

Explanation:
1. Masses give a factor 2 x 2 = 4.
2. Distance doubled gives a factor 1/4.
3. Net factor = 1, so the force is unchanged.

Teacher's Note:
1. Combine the factors from numerator and denominator.

Answer: (a) 19.6 m

Explanation:
1. \( H = u^2/2g = 19.6^2 / (2 \times 9.8) \).
2. = 384.16 / 19.6 = 19.6 m.

Teacher's Note:
1. At the top the speed is zero but acceleration is still g.

Answer: (c) ellipses with the sun at one of the foci

Explanation:
1. Kepler's first law: each planet moves in an elliptical orbit with the sun at one focus.

Teacher's Note:
1. The orbits of most planets are nearly circular.

Answer: (b) the gravitational pull of the earth

Explanation:
1. In orbit the engine is off.
2. Gravity of the earth provides the required centripetal force.

Teacher's Note:
1. The satellite is in continuous free fall around the earth.

Answer: (a) decreases

Explanation:
1. \( g' = GM/(R+h)^2 \).
2. As h increases the denominator increases, so g decreases.

Teacher's Note:
1. g falls off with the square of the distance from the centre.

Answer: (b) 11.2 km/s

Explanation:
1. \( v_e = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6.4\times10^6} \approx 11.2 \) km/s.

Teacher's Note:
1. Escape speed is \( \sqrt{2} \) times the near-earth orbital speed.

Answer: (c) the masses involved are enormous

Explanation:
1. \( F = Gm_1m_2/r^2 \) grows with the masses.
2. For stars and planets the masses are so huge that the force becomes large.

Teacher's Note:
1. For everyday objects G makes the force negligible.

Answer: (a) 1.6 m/s\( ^2 \)

Explanation:
1. \( g_m \approx g_e/6 = 9.8/6 \approx 1.6 \) m/s\( ^2 \).

Teacher's Note:
1. 26 m/s\( ^2 \) is roughly the value on Jupiter.

Answer: (d) the astronaut and the satellite are both in free fall with the same acceleration

Explanation:
1. Gravity at orbital height is still strong.
2. Astronaut and satellite fall together with the same acceleration, so there is no contact force between them.

Teacher's Note:
1. Weightlessness means no normal force, not no gravity.

Answer: (a) \( g = \dfrac{GM}{R^2} \)

Explanation:
1. Weight of mass m: \( mg = GMm/R^2 \).
2. Cancelling m gives \( g = GM/R^2 \).

Teacher's Note:
1. g does not depend on the mass of the falling body.

Answer: (c) a polar satellite

Explanation:
1. Polar satellites go around the earth in a north-south direction at about 500 to 800 km height.
2. Their period is about 100 minutes.

Teacher's Note:
1. They pass over every latitude several times a day.