Gravitation

Question 1

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

Accepted Answer

88 earth days. Kepler's third law gives T² proportional to R³, so a radius ratio of 4 makes the period ratio 4^(3/2) = 8. Mercury's year is therefore about 687/8 ≈ 86 days, closest to the 88 days it actually takes.

Answer

225 earth days

Answer

172 earth days

Answer

124 earth days

Question 2

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

Accepted Answer

27 N. Above the surface gravity follows the inverse-square law, g_h = g[R/(R + h)]². With h = R/3 the bracket is 3/4, so the weight falls by a factor of 9/16, giving 48 x 9/16 = 27 N. The linear approximation g(1 − 2h/R) is not valid at this height.

Answer

16 N

Answer

32 N

Answer

36 N

Question 3

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to:

Accepted Answer

R^(3/2). Kepler's third law states that the square of the orbital period is proportional to the cube of the orbital radius, T² ∝ R³. Taking the square root gives T ∝ R^(3/2).

Answer

R^(1/2)

Answer

R²

Answer

R³

Q1.The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

Q2.A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

Q3.In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to: