Gravitation — Set 4
Physics · गुरुत्वाकर्षण · Questions 31–40 of 60
The ratio of escape velocity to the circular orbital velocity near a planet’s surface is?
Correct Answer: C. sqrt(2)
Orbital speed near surface is v = sqrt(GM/R). Escape speed is v_e = sqrt(2GM/R). Therefore v_e/v = sqrt(2).
If the mass of a planet is doubled while its radius remains the same, its escape velocity becomes?
Correct Answer: A. sqrt(2) times
Escape velocity is v_e = sqrt(2GM/R). If M becomes 2M, v_e becomes sqrt(2) times larger. Radius staying the same keeps the denominator unchanged.
A 60 kg person stands on the Moon where g ≈ 1.62 m/s^2. What is the person’s weight there (approximately)?
Correct Answer: B. 97 N
Weight is W = mg. So W = 60×1.62 ≈ 97.2 N, which is about 97 N. The mass stays 60 kg, but weight changes with g.
A 1 kg mass at Earth’s surface has gravitational force roughly equal to?
Correct Answer: A. 9.8 N
Near Earth’s surface, gravitational force equals weight. For m = 1 kg, W = mg = 1×9.8 = 9.8 N. This is why g is often called field strength in N/kg.
Gravitational potential at a point is defined as?
Correct Answer: C. Work done per unit mass in bringing a test mass from infinity to that point
Potential is the work done per unit mass in bringing a test mass from infinity. The path does not matter because gravity is conservative. For a point mass, this gives V = -GM/r.
The value of the universal gravitational constant G in SI units is approximately?
Correct Answer: B. 6.67×10^-11
G is a universal constant in Newton’s law of gravitation. Its SI value is about 6.67×10^-11. Its unit is N m^2 kg^-2.
A geostationary satellite appears fixed over Earth because its orbital period is?
Correct Answer: D. Equal to Earth’s rotational period (about 24 hours)
To stay above the same point, the satellite must match Earth’s rotation. So its period must be about 24 hours. It must also orbit in the equatorial plane.
For a uniform solid sphere of density ρ and radius R, the surface gravity g is?
Correct Answer: C. (4/3)πGρR
For a uniform sphere, mass M = (4/3)πR^3ρ. Substituting into g = GM/R^2 gives g = (4/3)πGρR. This shows surface gravity increases with radius if density is fixed.
If the acceleration due to gravity g decreases, the time period of a simple pendulum (same length) will?
Correct Answer: D. Increase
The correct answer is 'Increase'. For a simple pendulum, T = 2π sqrt(l/g). If g becomes smaller, T becomes larger. That is why pendulum clocks run slow at higher altitudes.
The total mechanical energy E of a satellite of mass m in a circular orbit of radius r around mass M is?
Correct Answer: B. −GMm/(2r)
In a circular orbit, kinetic energy is GMm/(2r) and potential energy is −GMm/r. Adding them gives E = −GMm/(2r). The negative sign indicates a bound orbit.