JEE Physics Gravitation MCQs Set 30

Answer: (c) \( \dfrac12\sqrt{\dfrac{Gm}{R}} \)

Explanation:
1. The particles are at opposite ends of a diameter, so their separation is 2R.
2. Gravitational force provides the centripetal force: \( \dfrac{Gm^2}{(2R)^2}=\dfrac{mv^2}{R} \).
3. \( v^2=\dfrac{Gm}{4R}\Rightarrow v=\dfrac12\sqrt{\dfrac{Gm}{R}} \).

Teacher's Note:
1. The distance between the particles is the diameter, but the radius of the path is R. Do not mix the two.

Answer: (b) \( \dfrac{R_e}{15} \)

Explanation:
1. Energy conservation: \( \tfrac12v^2-\dfrac{GM}{R_e}=-\dfrac{GM}{R_e+h} \) with \( v^2=v_e^2/16=\dfrac{GM}{8R_e} \).
2. \( \dfrac{GM}{R_e}\left(\dfrac1{16}-1\right)=-\dfrac{GM}{R_e+h}\Rightarrow R_e+h=\dfrac{16}{15}R_e \).
3. So \( h=R_e/15 \).

Teacher's Note:
1. Do not use mgh at such heights. Use the potential energy formula minus GM over r.

Answer: (c) (a) and (c)

Explanation:
1. Outside the earth \( g\propto1/r^2 \): (a) true. Inside, \( g\propto r \), so g increases when moving away from the centre: (b) false.
2. At the centre g = 0: (c) true.
3. If rotation stops, the centrifugal effect disappears and g increases at the equator: (d) false.

Teacher's Note:
1. Inside the earth g is proportional to the distance from the centre. Rotation reduces the apparent g except at the poles.

Answer: (b) \( -4.8\times10^7 \)

Explanation:
1. Outside the earth \( g=GM/r^2 \) and \( V=-GM/r \), so \( V=-gr \).
2. \( V=-6\times8\times10^6=-4.8\times10^7\ \text{J/kg} \).

Teacher's Note:
1. Field times distance gives the magnitude of the potential for a spherically symmetric source. The potential is negative.

Answer: (d) (a), (b), (c) and (d)

Explanation:
1. To stay fixed over one point it must have the earth's period (24 h) and the same sense of rotation: west to east.
2. This needs an equatorial circular orbit of radius about 42000 km, i.e. about 36000 km above the surface. All four conditions are essential.

Teacher's Note:
1. A geostationary satellite has to match the earth's rotation in direction, period and plane. Only a circular equatorial orbit does that.

Answer: (b) move tangentially to the original orbit with the same velocity

Explanation:
1. With no force, Newton's first law applies: the satellite keeps its instantaneous velocity.
2. The velocity is tangent to the orbit, so it moves along a straight tangent with the same speed.

Teacher's Note:
1. Centripetal force is what bends the path into a circle. Removing it gives straight line motion along the tangent.

Answer: (c) \( mgR \)

Explanation:
1. Total energy at infinity must be zero: \( K-\dfrac{GMm}{R}=0 \).
2. \( K=\dfrac{GMm}{R}=mgR \) since \( GM=gR^2 \).

Teacher's Note:
1. Escape energy equals the magnitude of the gravitational potential energy on the surface, mgR.

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

Explanation:
1. At small height: \( \Delta g=\dfrac{2h}{R}g \). At small depth: \( \Delta g=\dfrac dRg \).
2. Equating, \( d=2h \).

Teacher's Note:
1. g falls twice as fast with height as with depth for small changes. Use the binomial approximation for both.

Answer: (c) \( 2R \)

Explanation:
1. \( g'=g\left(\dfrac{R}{R+h}\right)^2=\dfrac g9\Rightarrow R+h=3R \).
2. So \( h=2R \).

Teacher's Note:
1. Take the square root of the factor 1/9 to get the ratio of distances from the centre. Then subtract R.

Answer: (b) \( -\dfrac{9Gm}{r} \)

Explanation:
1. Let the null point be at distance x from m: \( \dfrac{Gm}{x^2}=\dfrac{4Gm}{(r-x)^2} \) gives \( r-x=2x \), so \( x=r/3 \).
2. Potential = \( -\dfrac{Gm}{r/3}-\dfrac{4Gm}{2r/3}=-\dfrac{3Gm}{r}-\dfrac{6Gm}{r}=-\dfrac{9Gm}{r} \).

Teacher's Note:
1. Field is a vector and cancels at the null point, but potential is a scalar and adds. The null point is nearer the smaller mass.

Answer: (c) F, 3F

Explanation:
1. The force between m and M does not depend on other masses, so it remains F.
2. The added mass 2m (in contact with m, essentially at the same place) attracts M with 2F. Total force = F + 2F = 3F.

Teacher's Note:
1. Gravitational forces are independent of each other and add as vectors. They do not get screened by nearby masses.

Answer: (a) \( R/4 \)

Explanation:
1. New g = 0.64g, so \( \left(\dfrac{R}{R+h}\right)^2=0.64\Rightarrow\dfrac{R}{R+h}=0.8 \).
2. \( R+h=1.25R\Rightarrow h=R/4 \).

Teacher's Note:
1. A 36 percent decrease leaves 0.64, whose square root is 0.8.

Answer: (d) \( -\dfrac{\pi GM}{L} \)

Explanation:
1. Radius of the semicircle: \( \pi r=L\Rightarrow r=L/\pi \).
2. Every element of the rod is at the same distance r from the centre, so \( V=-\dfrac{GM}{r}=-\dfrac{\pi GM}{L} \).

Teacher's Note:
1. All mass elements are equidistant from the centre, so the potential is just minus GM over r. Find r from the arc length.

Answer: (a) At the point nearest to the sun (perihelion)

Explanation:
1. Angular momentum is conserved: \( mvr=\text{constant} \), so speed is greatest where r is smallest.
2. Hence kinetic energy is maximum at the perihelion.

Teacher's Note:
1. A planet moves fastest when closest to the sun. This is Kepler's second law.

Answer: (c) decreases

Explanation:
1. At the equator the earth's radius is slightly larger and the centrifugal effect of rotation is greatest, so the effective g is smaller: \( g'=g-\omega^2R \).
2. Therefore the weight decreases.

Teacher's Note:
1. Mass stays the same but weight depends on the effective g. Rotation reduces g most at the equator.

Answer: (c) \( mgR\dfrac{n}{n+1} \)

Explanation:
1. \( \Delta U=-\dfrac{GMm}{R+nR}+\dfrac{GMm}{R}=\dfrac{GMm}{R}\left(1-\dfrac1{n+1}\right) \).
2. With \( GM=gR^2 \): \( \Delta U=mgR\dfrac{n}{n+1} \).

Teacher's Note:
1. For n very small this reduces to mgh. For n very large it tends to mgR.

Answer: (c) angular momentum

Explanation:
1. The gravitational force of the sun on a planet is central, so the torque about the sun is zero.
2. Angular momentum \( mvr \) is conserved, so speed varies inversely with distance from the sun.

Teacher's Note:
1. Central forces conserve angular momentum. Linear momentum of the planet is not conserved because the sun pulls on it.

Answer: (c) 6v

Explanation:
1. \( v_{orbit}=\sqrt{GM/r}\propto1/\sqrt r \).
2. \( \dfrac{v_B}{v_A}=\sqrt{\dfrac{4R}{R}}=2\Rightarrow v_B=2\times3v=6v \).

Teacher's Note:
1. Orbital speed falls as the inverse square root of the radius. A satellite at a quarter of the radius moves twice as fast.