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Find g at a height of 480 km above the Earth, and at a depth of 2400 km below its surface, treating…

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

Find $g$ at a height of $480\ \text{km}$ above the Earth, and at a depth of $2400\ \text{km}$ below its surface, treating the Earth as uniform. ( $g = 9.8\ \text{m/s}^2$ , $R = 6400\ \text{km}$ .)

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

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