JEE Physics Gravitation MCQs Set 09

Answer: (c) \(\sqrt{2} : 1\)

Explanation:
1. Orbital radii are \(r_A = 2R\) and \(r_B = 4R\).
2. Orbital speed \(v = \sqrt{\frac{GM}{r}}\), so \(p = mv \propto \frac{1}{\sqrt{r}}\) for equal masses.
3. \(\frac{p_A}{p_B} = \sqrt{\frac{4R}{2R}} = \sqrt{2} : 1\).

Teacher's Note:
1. Measure the orbital radius from the centre of the earth, not from the surface.
2. For equal masses, momentum is proportional to speed.

Answer: (d) \(g_1 : g_2 = R_1\rho_1 : R_2\rho_2\)

Explanation:
1. \(g = \frac{GM}{R^2}\) with \(M = \frac{4}{3}\pi R^3\rho\).
2. So \(g = \frac{4}{3}\pi G\rho R\), which is proportional to \(\rho R\).
3. Therefore \(g_1 : g_2 = R_1\rho_1 : R_2\rho_2\).

Teacher's Note:
1. Substitute mass in terms of density before comparing planets.
2. At the surface, \(g \propto \rho R\).

Answer: (c) \(\Delta T = \frac{3}{2}T\frac{\Delta R}{R}\)

Explanation:
1. Kepler's third law gives \(T^2 \propto R^3\), so \(T \propto R^{3/2}\).
2. Differentiate: \(\frac{\Delta T}{T} = \frac{3}{2}\frac{\Delta R}{R}\).
3. So \(\Delta T = \frac{3}{2}T\frac{\Delta R}{R}\).

Teacher's Note:
1. For small changes use the logarithmic derivative: the fractional change in \(T\) is the power times the fractional change in \(R\).
2. Here the power is \(\frac{3}{2}\).

Answer: (b) \(-\frac{mg_0R^2}{2(R+h)}\)

Explanation:
1. Total energy of a satellite is \(E = -\frac{GMm}{2r}\) with \(r = R + h\).
2. Using \(GM = g_0R^2\), \(E = -\frac{mg_0R^2}{2(R+h)}\).

Teacher's Note:
1. Total energy of a bound circular orbit is half of the potential energy and is negative.
2. Replace \(GM\) by \(g_0R^2\) to express it in \(g_0\).

Answer: (a) \(8\ \text{N/kg}\)

Explanation:
1. The field is perpendicular to the equipotential lines and has magnitude \(E = \frac{\Delta V}{\Delta d}\), where \(\Delta d\) is the perpendicular distance between neighbouring lines.
2. The lines are \(1\ \text{m}\) apart along the axis and inclined at \(30^\circ\), so \(\Delta d = 1\times\sin 30^\circ = 0.5\ \text{m}\).
3. \(E = \frac{4}{0.5} = 8\ \text{N/kg}\).

Teacher's Note:
1. The gravitational field points perpendicular to equipotential surfaces.
2. Use the perpendicular spacing, not the spacing along the axis.

Answer: (b) \(\frac{16R}{7}\)

Explanation:
1. Conservation of energy: \(\frac{1}{2}v^2 - \frac{GM}{R} = -\frac{GM}{r}\) at the highest point.
2. \(\frac{1}{2}\cdot\frac{9GM}{8R} - \frac{GM}{R} = \frac{9GM}{16R} - \frac{16GM}{16R} = -\frac{7GM}{16R}\).
3. Hence \(\frac{GM}{r} = \frac{7GM}{16R}\), so \(r = \frac{16R}{7}\).

Teacher's Note:
1. Write energy per unit mass and set the speed to zero at the top.
2. Keep the distance measured from the centre.

Answer: (c) \(\frac{g}{2}\)

Explanation:
1. Inside a uniform earth, \(g_d = g\left(1 - \frac{d}{R}\right)\).
2. With \(d = \frac{R}{2}\), \(g_d = g\left(1 - \frac{1}{2}\right)\).
3. So \(g_d = \frac{g}{2}\).

Teacher's Note:
1. Below the surface, \(g\) decreases linearly with depth to zero at the centre.
2. Above the surface it decreases as the inverse square.

Answer: (b) the mass of the projectile

Explanation:
1. \(v_e = \sqrt{\frac{2GM}{r}}\), which depends on \(G\), the mass \(M\) of the earth and the distance \(r\) of the point of projection from its centre.
2. The projectile's mass cancels out of the energy equation.
3. So it is independent of the mass of the projectile.

Teacher's Note:
1. Escape velocity is the same for a pebble and a rocket launched from the same place.
2. It also does not depend on the direction of projection.

Answer: (c) \(1 : 1\)

Explanation:
1. For a circular orbit, \(\frac{GMm}{r^2} = m\omega^2 r\), so \(\omega^2 = \frac{GM_{planet}}{r^3}\).
2. The satellite's own mass cancels.
3. Both have the same frequency, so the ratio is \(1 : 1\).

Teacher's Note:
1. The orbital period depends on the planet's mass and the radius of the orbit only.
2. The mass of the satellite never appears.

Answer: (a) \(\frac{k_0}{9}\)

Explanation:
1. Angular momentum is conserved: \(v_1 d = v_2(3d)\), so \(v_2 = \frac{v_1}{3}\).
2. Kinetic energy is proportional to \(v^2\), so it becomes \(\frac{1}{9}\) of the original.
3. \(KE_{far} = \frac{k_0}{9}\).

Teacher's Note:
1. At the extreme points the velocity is perpendicular to the radius, so \(mvr\) is constant.
2. The comet is slowest at the farthest point.

Answer: (b) \(9R\)

Explanation:
1. \(\frac{g'}{g} = \left(\frac{R}{R+h}\right)^2 = \frac{1}{100}\).
2. So \(\frac{R}{R+h} = \frac{1}{10}\), which gives \(R + h = 10R\).
3. \(h = 9R\).

Teacher's Note:
1. The distance from the centre is \(R + h\), not \(h\).
2. Take square roots on both sides of the ratio.

Answer: (b) \(\sqrt{ab}\)

Explanation:
1. \(v_e = \sqrt{2gR}\).
2. So \(\frac{v_1}{v_2} = \sqrt{\frac{g_1R_1}{g_2R_2}}\).
3. \(= \sqrt{ba}\).

Teacher's Note:
1. Escape velocity can be written as \(\sqrt{2gR}\) using \(GM = gR^2\).
2. The ratio follows without needing the masses.

Answer: (a) \(v_e\sqrt{1 - \eta^2}\)

Explanation:
1. Let the orbit radius be \(r\) and orbital speed \(v\). Energy conservation: \(\frac{1}{2}\eta^2v_e^2 - \frac{GM}{R} = \frac{1}{2}v^2 - \frac{GM}{r}\) per unit mass.
2. For a circular orbit \(v^2 = \frac{GM}{r}\), so the right side is \(-\frac{v^2}{2}\). With \(v_e^2 = \frac{2GM}{R}\), the left side is \(\frac{1}{2}v_e^2(\eta^2 - 1)\).
3. Hence \(v^2 = v_e^2(1 - \eta^2)\) and \(v = v_e\sqrt{1 - \eta^2}\).

Teacher's Note:
1. Circular-orbit condition \(v^2 = \frac{GM}{r}\) makes total energy \(-\frac{v^2}{2}\) per unit mass.
2. The orbit radius is \(r = \frac{R}{2(1-\eta^2)}\), which is at least \(R\) only when \(\eta \ge \frac{1}{\sqrt{2}}\).

Answer: (a) \(6\sqrt{2}\ \text{hour}\)

Explanation:
1. The orbital radii are \(7R\) and \(3.5R\).
2. \(T^2 \propto r^3\), so \(T = 24\left(\frac{3.5}{7}\right)^{3/2} = 24\left(\frac{1}{2}\right)^{3/2}\).
3. \(T = \frac{24}{2\sqrt{2}} = 6\sqrt{2}\ \text{hour}\).

Teacher's Note:
1. A geostationary satellite has a period of exactly one day.
2. Use Kepler's third law with orbital radii from the centre of the earth.

Answer: (a) \(1\)

Explanation:
1. A parking orbit is a geostationary orbit, with a period of \(24\ \text{h}\) and a fixed radius.
2. The period does not depend on the satellite's mass.
3. Both periods are equal, so the ratio is \(1\).

Teacher's Note:
1. All geostationary satellites share the same period and radius.
2. Mass does not matter for orbital motion.

Answer: (b) \(\frac{1}{6}\)

Explanation:
1. The field is maximum at the surface, \(E_s = \frac{GM}{R^2}\), so \(M = \frac{E_sR^2}{G}\).
2. From the graph, \(E_1 = 2\) at \(R_1 = 1\) and \(E_2 = 3\) at \(R_2 = 2\).
3. \(\frac{M_1}{M_2} = \frac{2\times 1^2}{3\times 2^2} = \frac{1}{6}\).

Teacher's Note:
1. The peak of the field graph is at the surface of the sphere.
2. Use \(E = \frac{GM}{R^2}\) at the surface to compare the masses.

Answer: (b) \(10\ \text{m}\)

Explanation:
1. The person gives the same take-off speed \(u\) in both places, so \(h = \frac{u^2}{2g}\) and \(h \propto \frac{1}{g}\).
2. \(h' = 2\times\frac{9.8}{1.96} = 2\times 5\).
3. \(h' = 10\ \text{m}\).

Teacher's Note:
1. For the same take-off speed, the jump height is inversely proportional to \(g\).
2. Use \(g = 9.8\ \text{m s}^{-2}\) on the earth.

Answer: (a) \(A\)

Explanation:
1. Angular momentum is conserved, so \(v r = \text{constant}\) at the extreme points.
2. The planet is nearest to the Sun at \(A\) (perihelion), so \(r\) is smallest and \(v\) is largest there.
3. Hence the speed is maximum at \(A\).

Teacher's Note:
1. Kepler's second law: the planet moves faster when it is closer to the Sun.
2. Perihelion is the fastest point and aphelion the slowest.

Answer: (b) \(\sqrt{\frac{3gh}{4}}\)

Explanation:
1. The block's mass is not negligible, so the earth also accelerates. The relative acceleration is \(g_{rel} = \frac{G(M_e + m)}{R^2} = g\left(1 + \frac{1}{3}\right) = \frac{4g}{3}\).
2. Relative speed after the separation falls by \(\frac{h}{2}\): \(v_{rel}^2 = 2\cdot\frac{4g}{3}\cdot\frac{h}{2} = \frac{4gh}{3}\).
3. The block's share of the relative speed is \(\frac{M_e}{M_e + m} = \frac{3}{4}\), so \(v^2 = \frac{9}{16}\cdot\frac{4gh}{3} = \frac{3gh}{4}\).

Teacher's Note:
1. When the falling body is as heavy as the earth, use the two-body relative motion.
2. Momentum conservation gives the block's share of the relative speed.

Answer: (d) \(\frac{2mgR}{3}\)

Explanation:
1. The final distance from the centre is \(3R\).
2. \(W = \Delta U = GMm\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2GMm}{3R}\).
3. With \(GM = gR^2\), \(W = \frac{2mgR}{3}\).

Teacher's Note:
1. Use \(GM = gR^2\) to convert the answer to \(g\) and \(R\).
2. Height \(2R\) means distance \(3R\) from the centre.

Answer: (c) \(\frac{GMm}{10R}\)

Explanation:
1. Total energy in the orbit is \(E = -\frac{GMm}{2r} = -\frac{GMm}{10R}\).
2. To escape, the total energy must become zero.
3. The minimum extra kinetic energy is \(\frac{GMm}{10R}\).

Teacher's Note:
1. Escape energy from an orbit equals the magnitude of the total energy, which is the same as the orbital kinetic energy.
2. Use the orbit radius \(5R\), not \(R\).

Answer: (c) areal velocity

Explanation:
1. Areal velocity is \(\frac{dA}{dt} = \frac{L}{2m}\), and \(L\) is constant for a central force.
2. For a circular orbit the angular velocity and energies are constant, but for an elliptical orbit they change.
3. Areal velocity stays constant in every orbit.

Teacher's Note:
1. Kepler's second law holds for any orbit, circular or elliptical.
2. Pick the quantity that is true in the general case.

Answer: (a) \(\sqrt{gR}\)

Explanation:
1. Starting from rest at distance \(2R\): \(\frac{1}{2}v^2 = GM\left(\frac{1}{R} - \frac{1}{2R}\right) = \frac{GM}{2R}\).
2. With \(GM = gR^2\), \(v^2 = gR\).
3. \(v = \sqrt{gR}\).

Teacher's Note:
1. Starting from rest means the initial kinetic energy is zero.
2. Use the energy difference between \(2R\) and \(R\).

Answer: (d) \(2\sqrt{\frac{2Gm}{l}}\)

Explanation:
1. Each fixed mass is at \(\frac{l}{2}\) from the mid-point, so the potential per unit mass there is \(V = -2\cdot\frac{Gm}{l/2} = -\frac{4Gm}{l}\).
2. For escape, \(\frac{1}{2}v^2 + V = 0\), so \(v^2 = \frac{8Gm}{l}\).
3. \(v = \sqrt{\frac{8Gm}{l}} = 2\sqrt{\frac{2Gm}{l}}\).

Teacher's Note:
1. Potentials add as scalars, so the two masses give twice the single-mass value.
2. Escape needs total energy zero.

Answer: (b) increase by \(2\%\)

Explanation:
1. \(g = \frac{GM}{R^2}\), so \(\frac{\Delta g}{g} = -2\frac{\Delta R}{R}\).
2. \(\frac{\Delta R}{R} = -1\%\), so \(\frac{\Delta g}{g} = +2\%\).
3. So \(g\) increases by about \(2\%\).

Teacher's Note:
1. A small percentage change in \(R\) changes \(g\) by twice that percentage, in the opposite direction.
2. Use the power of \(R\) in the formula.