Question

Your spaceship lands on an unknown planet. To determine the characteristics of this planet, you drop a wrench from 3.50 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

Step-by-step explanation:

By the formula,

`s=ut+\frac12at^2`

`-3.5=0+\frac12a(0.809)^2`

`a=-10.659 ms^(-2)`

where a is the acceleration of objects by gravity.

We also know that by the Law of Gravitation,

`F=-(GMm)/(r^2)=ma`

`a=-(GM)/(r^2)`

The mass of the planet is given by

`M=\rho V=\rho(\frac 43\pi r^3)`

So

`a=-(G(\rho\frac 43\pi R^3))/(r^2)= - ((6.67 * 10^(-11))(5500)(\frac43)\pi R^3)/((R+3.5)^2) = -10.659`

Since R>>3.5, so we approximate

`(R+3.5)^2\approx R^2`

Solving the last equation,

`R=6936.483 km`