On Mars, g = 3.7 m/s² and the radius is 3.39 × 10⁶ m…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

On Mars, $g = 3.7\ \text{m/s}^2$ and the radius is $3.39 \times 10^{6}\ \text{m}$ . Find its mass and its mean density. ( $G = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ .)

Need a reminder? The key ideaShow

At a planet's surface $g = \dfrac{GM}{R^2}$ . At a height $h$ : $g\left(\dfrac{R}{R + h}\right)^2$ . At a depth $d$ in a uniform planet: $g\left(1 - \dfrac{d}{R}\right)$ . A spinning planet's equator needs $\omega^2R$ of it for the circular motion.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← Jupiter turns once in 9.93 hours, its equator's radius is 7.15 × 10⁷ m, and its pull there gives…Taking R = 6400 km and a uniform Earth: At what height is g a sixteenth of its surface value? … →

More onMechanics › Gravitation Astrophysics

Discussion (0)

Loading…