Question #168538

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?

1
Expert's answer
2021-03-04T15:18:51-0500

a) the gravitational potential energy at the distance r is


E=GMmr=43πR3ρGmrE=-\frac{GMm}{r}=-\frac{\frac{4}{3}\pi R^3\rho Gm}{r}

If r = R + h,


E=43πR3ρGmR+hE=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+h}

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


E=43πR3ρGmR+0E=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+0}

The total energy is


K+E=0.5mv243πR2ρGm=43πR3ρGmR+hK+E=0.5mv^2-\frac{4}{3}\pi R^2\rho Gm=-\frac{\frac{4}{3}\pi R^3\rho Gm}{R+h}

G=3v28πρR2(1RR+h)1G=\frac{3v^2}{8\pi \rho R^2}\left(1-\frac{R}{R+h}\right)^{-1}

c)


G=3(303.6)28π(3340)(1.74106)2(11.741061.74106+21.5)1=6.641011Nm2kg2G=\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}


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