Question

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 4.50 m above the ground and measure that it hits the ground 0.809 s later. Assuming that the planet has the same density as that of earth (5500 kg/m3), what is the radius of the planet?

Answer 1

$8.98×10^6\:\text{m}$

Explanation:

First we need to find the acceleration due to gravity on the planet. The wrench took 0.809 s to fall from a height of 4.50 m so we can use the equation

$y = -\frac{1}{2}gt^2$

Solving for g, we get

$g = -\dfrac{2y}{t^2} = -\dfrac{2(-4.50\:\text{m})}{(0.809\:\text{s})} = 13.8\:\text{m/s}^2$

Recall that the acceleration due to gravity on a planet's surface can be written as

$g = G\dfrac{M_p}{R_p^2}$

We can express the mass of the planet $M_p$ in terms of its density $\rho$ as follows:

$M_p = \rho \left(\dfrac{4\pi}{3}R_p^3\right) = \dfrac{4\pi}{3}\rho R_p^3$

The expression for g then becomes

$g = \dfrac{G}{R_p^2} \left(\dfrac{4\pi}{3}\rho R_p^3\right) = \dfrac{4\pi G}{3}\rho R_p$

Solving for $R_p,$ we get

$R_p = \dfrac{3g}{4\pi G\rho}$

$\:\:\:\:\:\:\:= \left[\dfrac{3(13.8\:\text{m/s}^2)}{4\pi (6.674×10^{-11}\:\text{Nm}^2\text{/kg}^2)(5500\:\text{kg/m}^3)}\right]$

$\:\:\:\:\:\:\:= 8.98×10^6\:\text{m}$