## JenniferSmart1 2 years ago 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?

1. JenniferSmart1

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

2. JenniferSmart1

@UnkleRhaukus

3. JenniferSmart1

@Jemurray3

4. Jemurray3

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

5. JenniferSmart1

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

6. JenniferSmart1

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

7. Jemurray3

I don't know what you mean by that.

8. JenniferSmart1

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

9. JenniferSmart1

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

10. Jemurray3

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

11. JenniferSmart1

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

12. JenniferSmart1

That worked, Thanks!