Gravitation — Practice

Question 1

Two masses of 10 kg and 20 kg are placed 1.0 m apart. Find the gravitational force between them.

Accepted Answer

F = G m1 m2 / r^2 = (6.67e-11 x 10 x 20) / 1^2 = 1.33e-8 N. The force is attractive and acts along the line joining them. Note how tiny it is for ordinary masses — this is why gravity is only noticeable when one body is planet-sized.

Question 2

Find g at a height equal to R above Earth's surface.

Accepted Answer

Distance from the centre = 2R, so g' = GM/(2R)^2 = GM/4R^2 = g/4 = 2.45 m/s^2. Use the exact inverse-square form here, not g(1 - 2h/R), because h is not small compared with R.

Question 3

Why is g zero at the centre of the Earth?

Accepted Answer

At the centre, every shell of the Earth has you inside it, and a uniform spherical shell exerts zero gravitational force on a mass anywhere inside it. With no mass left beneath you to pull, the net force and hence g are zero.

Question 4

At what depth below the Earth's surface does g fall to half its surface value? Take R = 6400 km.

Accepted Answer

Below the surface g_d = g(1 - d/R). Setting g_d = g/2 gives 1 - d/R = 1/2, so d = R/2 = 3200 km. Note the contrast: g halves at a depth of R/2, but going up it takes a height of about 0.41R to halve, because the two variations follow different laws.

Question 5

Find the orbital velocity of a satellite at 300 km height (R = 6400 km, M = 6e24 kg).

Accepted Answer

Orbital radius r = 6400 + 300 = 6700 km = 6.7e6 m. v_o = sqrt(GM/r) = sqrt(6.67e-11 x 6e24 / 6.7e6) = sqrt(5.97e7) = 7.73 km/s. The satellite's own mass never enters — it cancels when gravity is set equal to the centripetal force.

Question 6

Calculate the escape velocity on the Moon (g = 1.62 m/s^2, R = 1740 km).

Accepted Answer

v_e = sqrt(2gR) = sqrt(2 x 1.62 x 1.74e6) = sqrt(5.64e6) = 2.37 km/s, roughly five times smaller than Earth's 11.2 km/s. This is why the Moon has no atmosphere: gas molecules routinely exceed 2.37 km/s and simply leak away into space.

Question 7

If Earth's period is 1 year at orbital radius R, find the period of a planet at radius 4R.

Accepted Answer

Kepler's third law gives T^2 proportional to a^3, so (T/T_E)^2 = (4R/R)^3 = 64, giving T/T_E = 8 and T = 8 years. The constant of proportionality is the same for every planet orbiting the same Sun, which is why neither mass is needed.

Question 8

Find the gravitational potential energy of a 100 kg satellite at a height of 3600 km above the Earth's surface (R = 6400 km, M = 6e24 kg).

Accepted Answer

Distance from the centre r = 6400 + 3600 = 10000 km = 1.0e7 m. U = -GMm/r = -(6.67e-11 x 6e24 x 100)/1.0e7 = -4.0e9 J. The value is negative because zero potential energy is taken at infinity, and the satellite is bound to the Earth.

Question 9

A 200 kg satellite orbits at a height of 400 km (R = 6400 km, M = 6e24 kg). Find its total energy, and the extra energy needed to make it escape.

Accepted Answer

Orbital radius r = 6.8e6 m and GM = 4.0e14. Total energy E = -GMm/2r = -(4.0e14 x 200)/(2 x 6.8e6) = -5.89e9 J. To escape, the satellite's total energy must be raised to zero, so the extra energy required is +5.89e9 J. Note this is less than launching the same 200 kg from rest on the ground, because the satellite already carries kinetic energy in a useful direction.

Question 10

A body is projected vertically from the Earth's surface with a speed equal to half the escape speed. Find the maximum height it reaches, in terms of the Earth's radius R.

Accepted Answer

Take v = v_e/2, so v^2 = (2GM/R)/4 = GM/2R. Energy conservation between the surface and the highest point, where the speed is zero: (1/2)m(GM/2R) - GMm/R = -GMm/(R+h). The left side is GMm/4R - GMm/R = -3GMm/4R, so 3/(4R) = 1/(R+h), giving R + h = 4R/3 and h = R/3, about 2130 km. Half the escape speed does not get you half way — energy goes as the square of the speed.

Practice — Gravitation

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