What is the Velocity of a Person at the Equator?

[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) ?

 

[nrc_8706]

Hello?!

Can Anybody Help Me?!

 

[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.

 

[nrc_8706]

...

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

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

 

[nrc_8706]

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

 

[andrewchang]

where is the person?

 

[nrc_8706]

equartor

he is at the equator

 

[andrewchang]

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

 

[nrc_8706]

at the Earth's equator

 

[andrewchang]

yeah, so use the Earth's equator..

 

[nrc_8706]

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

 

[daveed]

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

 

[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?

 

[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?

 

[andrewchang]

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?

there should be an "r"

v=((GM/r^2 - g)*r)^1/2more importantly, do you understand how i got that?

 

[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.