JEE Physics Gravitation MCQs Set 13

Answer: (b) \(-mv^2\)

Explanation:
1. For a circular orbit \(\frac{GMm}{r} = mv^2\).
2. The potential energy is \(-\frac{GMm}{r} = -mv^2\).

Teacher's Note:
1. Potential energy is minus twice the kinetic energy.
2. GMm over r equals mv squared.

Answer: (b) 9.144 km/s

Explanation:
1. The orbit radius is \(3R\). Energy conservation gives \(\frac12v^2 = GM\left(\frac1R - \frac{1}{3R}\right) = \frac{2GM}{3R}\).
2. So \(v^2 = \frac23\cdot\frac{2GM}{R} = \frac23v_e^2\) and \(v = 11.2\sqrt{\frac23}\approx9.14\) km/s.

Teacher's Note:
1. Distance from the centre is height plus R, i.e. 3R.
2. Use the exact potential.

Answer: (d) r will decrease and v will increase

Explanation:
1. With less energy the satellite can no longer stay in the larger orbit, so it spirals into a smaller radius.
2. For circular orbits \(v = \sqrt{\frac{GM}{r}}\), so a smaller r gives a larger speed (the lost energy comes from the decrease in potential energy).

Teacher's Note:
1. Satellites speed up when they drop to a lower orbit.
2. Total energy is more negative in a lower orbit.

Answer: (a) \(-0.5\,mgR_e\)

Explanation:
1. The height equals \(R_e\), so the distance from the centre is \(2R_e\).
2. \(U = -\frac{GMm}{2R_e} = -\frac{gR_e^2m}{2R_e} = -0.5\,mgR_e\).

Teacher's Note:
1. Add the radius to the height first.
2. GM equals gR squared.

Answer: (a) \(E_k\)

Explanation:
1. In a circular orbit the total energy is \(-E_k\).
2. To escape, the total energy must become zero, so an extra kinetic energy \(E_k\) is needed.

Teacher's Note:
1. Total energy of orbit is minus its kinetic energy.
2. Escape needs zero total energy.

Answer: (a) positive

Explanation:
1. The total energy is \(\frac12mv^2 - \frac{GMm}{R}\).
2. Escape velocity makes this exactly zero, so any greater speed makes it positive.
3. A positive total energy means the missile leaves the earth with speed to spare.

Teacher's Note:
1. Total energy greater than zero means an unbound path.
2. Zero is the borderline for escape.

Answer: (d) \(4E_0\)

Explanation:
1. For a circular orbit \(U = 2E\) where E is the total energy.
2. With \(E = 2E_0\), \(U = 4E_0\) (here \(E_0\) itself is a negative number).

Teacher's Note:
1. Potential energy is twice the total energy in a circular orbit.
2. Kinetic energy equals minus the total.

Answer: (c) conservation of angular momentum

Explanation:
1. The gravitational force on a planet is central, so its torque about the Sun is zero.
2. Hence angular momentum is conserved, which means the radius vector sweeps out equal areas in equal times.

Teacher's Note:
1. Areal velocity is L over 2m.
2. Central force means constant angular momentum.

Answer: (c) at C

Explanation:
1. By conservation of angular momentum, \(v\) is largest at the nearest point (perihelion, A) and smallest at the farthest point (aphelion, C).
2. The Sun is near A, so C is the farthest point and the speed is minimum there.

Teacher's Note:
1. Speed is least when the planet is farthest from the Sun.
2. Perihelion has maximum speed.

Answer: (b) 60

Explanation:
1. Angular momentum is conserved: \(v_{near}r_{near} = v_{far}r_{far}\).
2. \(v_{near} = 12\times\frac{8\times10^{12}}{1.6\times10^{12}} = 12\times5 = 60\) m/s.

Teacher's Note:
1. Product of speed and distance is constant at the extreme points.
2. The comet is fastest near the sun.

Answer: (c) angular momentum

Explanation:
1. The gravitational force is central, so the torque about the earth's centre is zero and angular momentum is conserved.
2. Velocity and angular velocity change along an elliptical orbit.

Teacher's Note:
1. Central force means angular momentum is constant.
2. Speed varies along an ellipse.

Answer: (a) \(\frac{vr}{R}\)

Explanation:
1. Angular momentum is conserved between the closest and farthest points, where the velocity is perpendicular to the radius.
2. \(vr = v'R\), so \(v' = \frac{vr}{R}\).

Teacher's Note:
1. At the extreme points \(L = mvr\).
2. Farther point means smaller speed.

Answer: (c) \(\sqrt{2Emr^2}\)

Explanation:
1. \(E = \frac12mv^2\), so \(v = \sqrt{\frac{2E}{m}}\).
2. \(L = mvr = mr\sqrt{\frac{2E}{m}} = \sqrt{2Em\,r^2}\).

Teacher's Note:
1. Write v in terms of E and substitute.
2. Bring m inside the root carefully.

Answer: (a) \(\frac{2mA}{T}\)

Explanation:
1. The areal velocity is \(\frac{dA}{dt} = \frac{L}{2m}\), which is constant.
2. In one period the area swept is A, so \(\frac AT = \frac{L}{2m}\).
3. Hence \(L = \frac{2mA}{T}\).

Teacher's Note:
1. Kepler's second law links angular momentum to areal velocity.
2. Use total area over total time.

Answer: (d) \((r_1 + r_2)^3\)

Explanation:
1. By Kepler's third law \(T^2\propto a^3\), where a is the semi-major axis.
2. Here \(a = \frac{r_1 + r_2}{2}\), so \(T^2\propto(r_1 + r_2)^3\).

Teacher's Note:
1. Use the semi-major axis in Kepler's third law.
2. Major axis equals the sum of the extreme distances.

Answer: (d) 8 years

Explanation:
1. \(T^2\propto r^3\), so \(T\propto r^{3/2}\).
2. \(T' = 1\times4^{3/2} = 8\) years.

Teacher's Note:
1. Period depends on distance to the power 3/2.
2. 4 to the power 3/2 is 8.

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

Explanation:
1. \(T^2 = \frac{4\pi^2a^3}{GM}\) with \(a = \frac{r_A + r_P}{2}\) gives \(T^2 = \frac{\pi^2}{2GM}(r_A + r_P)^3\), so (b) is correct and (a) is not.
2. Angular momentum conservation gives \(v_Ar_A = v_Pr_P\), so (c) is correct.
3. Since \(r_A > r_P\), \(v_A < v_P\), so (d) is correct. Hence (b), (c) and (d).

Teacher's Note:
1. Check each statement separately.
2. Semi-major axis is half the sum of the extreme distances.

Answer: (b) \(\left(\frac{gR^2}{r^3}\right)^{1/2}\)

Explanation:
1. Gravity gives the centripetal force: \(\frac{GMm}{r^2} = m\omega^2r\), so \(\omega^2 = \frac{GM}{r^3}\).
2. With \(GM = gR^2\), \(\omega = \left(\frac{gR^2}{r^3}\right)^{1/2}\).

Teacher's Note:
1. Replace GM by gR squared.
2. Orbit radius appears cubed.

Answer: (d) \(\frac1R\)

Explanation:
1. Centripetal force: \(\frac{mv^2}{R}\propto\frac{1}{R^3}\).
2. So \(v^2\propto\frac{1}{R^2}\) and \(v\propto\frac1R\).

Teacher's Note:
1. Equate the force law to mv squared over R.
2. The exponent changes with the force law.

Answer: (d) S1 and S2 are moving with the same speed.

Explanation:
1. Orbital speed \(\sqrt{\frac{GM}{r}}\) and period \(2\pi\sqrt{\frac{r^3}{GM}}\) do not depend on the satellite's mass, so (b) is false and (d) is true.
2. Kinetic energy \(\frac12mv^2\) and potential energy \(-\frac{GMm}{r}\) are proportional to the mass, so (a) and (c) are false.

Teacher's Note:
1. Speed and period are independent of the satellite mass.
2. Energies are proportional to the mass.

Answer: (d) \(t_1 = 2t_2\)

Explanation:
1. By Kepler's second law the area swept per unit time is constant.
2. So the time is proportional to the area swept, and twice the area needs twice the time: \(t_1 = 2t_2\).

Teacher's Note:
1. Equal areas in equal times.
2. Time is proportional to area.

Answer: (c) 6V

Explanation:
1. \(v\propto\frac{1}{\sqrt r}\), so \(\frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt4 = 2\).
2. So \(v_B = 6V\).

Teacher's Note:
1. Nearer satellite moves faster.
2. Take the square root of the radius ratio.

Answer: (b) \(-\frac{3GM}{a}\)

Explanation:
1. The point is inside the shell, where the potential due to the shell is constant: \(-\frac{GM}{a}\).
2. The particle at the centre gives \(-\frac{GM}{a/2} = -\frac{2GM}{a}\).
3. The total is \(-\frac{3GM}{a}\).

Teacher's Note:
1. Potential inside a shell is constant and equal to its surface value.
2. Add scalar contributions.

Answer: (a) \(6\sqrt2\)

Explanation:
1. The geostationary satellite has \(r_1 = 6R\) and \(T_1 = 24\) h; the other satellite has \(r_2 = 3R\).
2. \(T\propto r^{3/2}\), so \(T_2 = 24\left(\frac{3R}{6R}\right)^{3/2} = \frac{24}{2\sqrt2} = 6\sqrt2\) h.

Teacher's Note:
1. Add R to each height to get the orbit radius.
2. Use Kepler's third law with the 24 hour satellite.

Answer: (a) 3R

Explanation:
1. \(\left(\frac{R}{R+h}\right)^2 = \frac{1}{16}\), so \(R + h = 4R\).
2. Hence \(h = 3R\).

Teacher's Note:
1. Take the square root of 1/16.
2. Subtract R from the distance from the centre.

Answer: (b) \(v_e = \sqrt2v_0\)

Explanation:
1. Orbital speed close to the surface is \(v_0 = \sqrt{gR}\) and the escape speed is \(v_e = \sqrt{2gR}\).
2. So \(v_e = \sqrt2v_0\).

Teacher's Note:
1. Escape speed is root two times the orbital speed.
2. This applies at the surface.