JEE Physics Gravitation MCQs Set 08

Answer: (a) 32 km

Explanation:
1. \(g'=g\left(\frac{R}{R+h}\right)^2\approx g\left(1-\frac{2h}{R}\right)\) for small \(h\)
2. A decrease of 1% gives \(\frac{2h}{R}=0.01\)
3. \(h=0.005\times6400=32\) km

Teacher's Note:
1. For heights much smaller than the radius use the binomial approximation. The exact calculation also gives about 32 km.

Answer: (c) \(\tfrac12mgR\)

Explanation:
1. \(W=U_f-U_i=-\frac{GMm}{2R}+\frac{GMm}{R}=\frac{GMm}{2R}\)
2. With \(GM=gR^2\) this is \(\frac{mgR}{2}\)

Teacher's Note:
1. Do not use \(mgh\) for large heights, because \(g\) is not constant. Use the potential energy \(-GMm/r\).

Answer: (a) 11 km/s

Explanation:
1. Escape velocity is found from energy: \(\tfrac12mv_e^2=\frac{GMm}{R}\)
2. Energy is a scalar, so the direction of projection does not matter
3. The escape velocity remains 11 km/s

Teacher's Note:
1. Escape velocity is independent of the mass of the body and of the direction of launch (if the path does not hit the planet).

Answer: (c) \(2^{-3/2}\) lunar month

Explanation:
1. Kepler's third law: \(T^2\propto r^3\)
2. \(\frac{T}{T_{moon}}=\left(\frac12\right)^{3/2}=2^{-3/2}\)

Teacher's Note:
1. Use ratios in Kepler's third law so that the constants cancel.

Answer: (c) \(d=2\) km

Explanation:
1. At height \(h\): \(g_h\approx g\left(1-\frac{2h}{R}\right)\); at depth \(d\): \(g_d=g\left(1-\frac dR\right)\)
2. Equating gives \(d=2h=2\) km

Teacher's Note:
1. Gravity falls twice as fast above the surface as it does below, for small distances.

Answer: (c) 1 Kinetic, 2 Total, 3 Potential

Explanation:
1. For a circular orbit \(K=+\frac{GMm}{2r}\), \(U=-\frac{GMm}{r}\) and \(E=-\frac{GMm}{2r}\)
2. Kinetic energy is the only positive curve, so curve 1 is kinetic
3. The potential energy is twice as negative as the total energy, so the lower curve 3 is potential and the upper negative curve 2 is total

Teacher's Note:
1. Remember the ratio K : E : U = 1 : -1 : -2 for circular orbits. The total energy is always negative for a bound orbit.