Question

At what distance \(h\) above the surface of the Earth is the free-fall acceleration half its value at sea level?

Here is what I did:
\[a=\frac{F_g}{m}=\frac{Gm_E}{r^2}\]
therefore....
\[r_2=\sqrt{\frac{Gm}{a/2}}\]
where \(r_2\) is the \(h\) that we're looking for.

What did I do wrong?

Answer 1

The answer I get is \(8.12\times 10^7m\) which is wrong :(

Answer 2

@UnkleRhaukus

Answer 3

@Jemurray3

Answer 4

You solved for the radial distance from the center of the earth -- you have to subtract the earth's radius from that value.

Answer 5

do you get 9.83E6 for a_1? or would I use 9.8 m/s^2?

Answer 6

I'm surprised that my a_1 doesn't equal 9.8, because that's what the acceleration at sea level is

Answer 7

I don't know what you mean by that.

Answer 8

Here is how I calculated the acceleration at sea level:
\[a=\frac{Gm_E}{r^2}\]
\[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}\]

Answer 9

\[a=\frac{6.67\times 10^{-11}Nm^2/kg^2\cdot 5.98\times10^{24}kg}{(6370m)^2}=9.83\times 10^6 m/s^2\]

Answer 10

6370 meters? That's about four miles... :)

Answer 11

LOL! OOOOooops!!! Let's try that again

Answer 12

That worked, Thanks!