Taking R = 6400 km and a uniform Earth: At what height is g a sixteenth of its surface value? …

Contributed by StemPy Lab

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

Taking $R = 6400\ \text{km}$ and a uniform Earth:

At what height is $g$ a sixteenth of its surface value?
At what depth is $g$ $0.4$ times its surface value?
At what height is $g$ $1\%$ smaller than at the surface?

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.

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