Answer to Question #182072 in Astronomy | Astrophysics for Afnan Ibn Kamal

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 p of 3340 kg/m^3 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:

E=(-4πG/3)mp.R^3/(R+H)

(b) Next, show that the gravitational constant can be determined by:

G=(3/8π)×(v^2/‌pR^2)×[1-{R/(R+H)}]^-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 potential energy of a piece with mass m at a distance r from the spherically symmetric object of mass M is

$E = -\dfrac{GMm}{r}$

If the Moon has radius R and density $\rho,$ then $M =\dfrac43\pi R^3\rho$ .

$E = -\dfrac{GMm}{r} = -\dfrac{\frac43 \pi G R^3\rho m}{r} = -\dfrac{\frac43 \pi G R^3\rho m}{R+H}$ .

b) If H is the maximum height, then the object stops there. Let us determine the total energy of the rock at the surface of the Moon and at the height h

$E_{tot,1} = \dfrac{mv^2}{2} - \dfrac{\frac43 \pi G R^3\rho m}{R}$ ,

$E_{tot,2} = 0 - \dfrac{\frac43 \pi G R^3\rho m}{R+H}$ .

Therefore,

$\dfrac{mv^2}{2} = \dfrac{\frac43 \pi G R^3\rho m}{R} - \dfrac{\frac43 \pi G R^3\rho m}{R+H} = \dfrac{\frac43 \pi G R^3\rho m}{R} \cdot \Big(1 - \dfrac{R}{R+H}\Big).$

$G = \dfrac{mv^2}{2\cdot \frac43\pi \rho m R^2 } \cdot \Big(1 - \dfrac{R}{R+H}\Big)^{-1}, \\ G = \dfrac{3v^2}{8\pi \rho R^2 } \cdot \Big(1 - \dfrac{R}{R+H}\Big)^{-1}.$

c) $G = \dfrac{3v^2}{8\pi \rho R^2 } \cdot \Big(1 - \dfrac{R}{R+H}\Big)^{-1} = \dfrac{3\cdot(30000/3600)^2}{8\pi\cdot 3340\cdot 1740000^2}\cdot \Big(1 - \dfrac{1740000}{1740000+21.5} \Big)^{-1} = 6.634\cdot10^{-11}\,\mathrm{N\cdot m^2/kg^2}$