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:

E = − 4πG 3 mρ · R3 R + h

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

G = 3 8π v 2 ρR2  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?

(PLEASE SOLVE ALL PARTS)

Answer 1

(a) the gravitational potential energy at the distance r is

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

If r = R + h,

$E = -\dfrac{G\cdot\frac43 \pi R^3m \rho}{R+h}$ .

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

$E = -\dfrac{G\cdot\frac43 \pi R^3m \rho}{R+0} = -\frac43\cdot G\pi R^2m \rho.$

The total energy is $K +E= \dfrac{mv^2}{2} -\frac43\cdot G\pi R^2m \rho.$

At the height h the kinetic energy is 0, so the total energy is $-\dfrac{\frac43 \cdot G \pi R^3m \rho}{R+h}$ .

According to the law of conservation of energy,

$\dfrac{mv^2}{2} -\frac43\cdot G\pi R^2m \rho = -\dfrac{\frac43 \cdot G \pi R^3m \rho}{R+h}$ ,

$\dfrac{mv^2}{2} = \frac43\cdot G\pi m \rho \left( \dfrac{R^3+R^2h-R^3}{R+h}\right), \\ v^2 = \frac83 \pi G\rho \dfrac{R^2h}{R+h}, \\ v^2 = \frac83 \pi G\rho R^2 \left( 1-\dfrac{R}{R+h} \right), \\ G = \frac38 \dfrac{v^2}{\pi \rho R^2} \left( 1-\dfrac{R}{R+h} \right)^{-1}.$

(c) Let us substitute all the parameters known

$G = \frac38 \dfrac{v^2}{\pi \rho R^2}\cdot \left( 1-\dfrac{R}{R+h} \right)^{-1} = \frac38 \cdot \dfrac{(30/3.6 \,\mathrm{m/s})^2}{3.14\cdot 3340\,\mathrm{kg/m^3} \cdot (1.74\cdot10^6\,\mathrm{m})^2}\cdot \left( 1-\dfrac{1.74\cdot10^6\,\mathrm{m}}{1.74\cdot10^6\,\mathrm{m}+ 21.5\,\mathrm{m}} \right)^{-1} = 6.64\cdot10^{-11}\,\mathrm{N/kg^2\cdot m^2} .$