Answer to Question #168538 in Astronomy | Astrophysics for Buthmini Sehansa

Question

An astronaut working on the Moon tries to determine the gravitational constant G by throwing a Moon rock of mass m with a velocity of v vertically into the sky. The astronaut knows that the Moon has a density ρ of 3340 kg/m3 and a radius R of 1740 km.(a) Show with (1) that the potential energy of the rock at height h above the surface is given by:
4πG R3
E = −
3 mρ · R + h (2)
(b) Next, show that the gravitational constant can be determined by:
3 v2 .
R Σ−1
(c) What is the resulting G if the rock is thrown with 30 km/h and reaches 21.5 m?

Answer 1

a) the gravitational potential energy at the distance r is

$$E=-\frac{GMm}{r}=-\frac{\frac{4}{3}\pi R^3\rho Gm}{r}$$

If r = R + h,

$$E=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+h}$$

b) At the surface of the Moon the potential energy is

$$E=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+0}$$

The total energy is

$$K+E=0.5mv^2-\frac{4}{3}\pi R^2\rho Gm=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+h}$$

$$G=\frac{3v^2}{8\pi \rho R^2}\left(1-\frac{R}{R+h}\right)^{-1}$$

c)

$$G=\frac{3(\frac{30}{3.6})^2}{8\pi (3340)(1.74⋅10 ^ 6)^2}\left(1-\frac{1.74⋅10 ^ 6}{1.74⋅10 ^ 6+21.5}\right)^{-1}\\=6.64\cdot10^{-11}\frac{Nm^2}{kg^2}$$