Question

I know the answer to this is question is g/4, I just wonder why it is that. Can someone explain? - What is the magnitude of g at an altitude equal to the radius of the earth?

Answer 1

Newton's law of gravitation tells us that:

\[F=G\frac{m_e m}{r^{2}}\]

where \(G\) is a constant, \(m_e\) is the mass of the Earth and \(m\) is your object while \(r\) is the distance between the centre of mass of the two objects.

Newton Second Law also has something to say about \(F\):
\[F=ma\]

Now since \(F=F\) (obviously!):

\[ma=G\frac{m_e m}{r^{2}}\]

Dividing both sides by \(m\) we get:
\[a=G\frac{m_e }{r^{2}}\]

Which is the acceleration of mass \(m\) at distance \(r\) from the centre of mass \(m_e\)

Answer 2

But we want to know what \(a\) is at a certain altitude compared to the surface of the earth:
\[\frac{a_h}{a}=\dfrac{G\frac{m_e}{r_h^2}}{G\frac{m_e}{r^2}}=\frac{r^2}{r_h^2}\]

Simplifing the fractions gives us:
\[\frac{a_h}{a}=\frac{r^2}{r_h^2}\]

Answer 3

At ground level we have acceleration \(a=g\), and distance from centre of earth as \(r=r_e\).

At height \( h\) we are \(r=2r_e\) from the centre of earth.

Plugging these in:
\[\frac{a_h}{a}=\frac{r^2}{r_h^2}\]

Gives
\[\frac{a_h}{g}=\frac{r_e^2}{(2r_e)^2}=\frac{1}{4}\]

And ultimately the answer you got:
\[a_{h}=\frac{g}{4}\]

:)