Centripetal acceleration

1. Dec 4, 2005

nrc_8706

Given: r(Earth)=6.37*10^6 m
m(earth)=5.98*10^24 kg
r(moon)=1.74*10^6m
g=9.8 m/s^2
G=6.67259*10^-11 Nm^2/kg^2

Calculate the speed of a 97.4kg person at the equator.
do u use a=v^2/r v=sqr root (a*r) ?

2. Dec 4, 2005

nrc_8706

Hello????!!!!!!

Can Anybody Help Me?!

3. Dec 4, 2005

andrewchang

$$\Sigma F = ma_c$$

... you know that gravity is the force acting towards the center, you know the radius, so you can find the v.

4. Dec 4, 2005

nrc_8706

.....

ok, taking what you said i got that Gravity=m*v^2/r

v=(gr/m)^1/2 correct?

5. Dec 4, 2005

nrc_8706

which radius do u use? the earth's or the moon's?

6. Dec 4, 2005

andrewchang

where is the person?

7. Dec 4, 2005

nrc_8706

equartor

he is at the equator

8. Dec 4, 2005

andrewchang

of which? the earth or the moon? which radius does it make sense to use?

9. Dec 4, 2005

nrc_8706

at the earth's equator

10. Dec 4, 2005

andrewchang

yeah, so use the earth's equator..

11. Dec 4, 2005

nrc_8706

do u use 9.8m/s^2 or 6.67*10^-11

12. Dec 4, 2005

daveed

those are two different things... you can use either, if you know what you're doing.

Last edited: Dec 4, 2005
13. Dec 4, 2005

andrewchang

$$F_g - N = m \times \frac {v^2}{r}$$
$$\frac{GMm}{r^2} - mg = \frac {mv^2}{r}$$
$$\frac{GM}{r^2} - g = \frac{v^2}{r}$$

now solve for v.

do you understand what i did?

Last edited: Dec 4, 2005
14. Dec 4, 2005

nrc_8706

v=(GM/r^2-g)^1/2

=(((6.67*10^-11*5.98*10^24)/(6.37*10^6)^2)-9.8)^1/2 correct?

15. Dec 4, 2005

andrewchang

there should be an "r"

v=((GM/r^2 - g)*r)^1/2

more importantly, do you understand how i got that?

Last edited: Dec 4, 2005
16. Dec 4, 2005

nrc_8706

yes thank you. i knew it had something to do with F=GMm/R^2 i just didnt know what to do. thank for all your help and your patience.

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