Gravatational Potential Energy

[Lemmy]

Okay I understand somewhat of this topic of grav potential energy but i got stumped on this one question. I am to find the mass of a satellite orbiting earth with the given variables of (F)(units: kN) which is the force the sattelite attracts the earth, and given the gravitational potential energy (-U). You are also givin the gravitational constant (G), and the earth (m_e)

How could you find the mass (m) of the sattelite without a given radius in the equation U=-(G*m_e*m)/r ?

How would i rewrite this equation

i know if i was givin radius then i could write it as m=(U*r)/-(G*m_e)

any help would be great thanks
lemmy

here is the problem
"When in orbit, a communication satellite attracts the earth with a force of F and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - U"
"Find the mass of the satellite."
"Take the gravitational constant to be G , the mass of the Earth to be m_e ."

[Lemmy]

Hmmm maybe F=(G*m*m_e)/r^2, i could possibly do r^2=(G*m*m_e)/F then r=sqrt((G*m*m_e)/F)
u think thats how to do radius?

[Lemmy]

tell me if this sounds right, m=(U*sqrt((G*m*m_e)/F))/(G*m_e)
aww nm i don't know m :(

[Physics Monkey]

Hi again Lemmy, please don't post multiple copies of the same question in different forums, thanks!

Regarding your work so far, you have found a formula for the radius in terms of the unknown mass and known quantities (m_e, G, F). Can you now find another formula that relates m and r so that you can plug in and solve for m?

[Lemmy]

well U= -(G*m_e*m)/(r) and F=-(G*m_e*m)/r^2 is all that comes to mind right now

[Physics Monkey]

Lemmy, you keep going in circles. I can't just tell you the answer, but it is staring you right in face. Think about what equations you have used already and what equations you haven't used yet. Hint: you know r in terms of m, so look back at your first post.

[Lemmy]

what do you mean i know r in terms of m? and i still don't understand where for would apply,tricky part is i'm not allowed to have r in my answer

[Lemmy]

i'm trying to find the equation to solve this problem

[Physics Monkey]

Lemmy, in post number 2 you give a formula for radius r in terms of mass m, G, m_e, and F. This formula came from the force equation. In post one you give a formula for mass m in terms of radius r, G, m_e, and U. This formula came from the potential energy equation. Can you somehow combine the two?

Edit: Look at post 3, you've already done it!

[Lemmy]

m=(U*sqrt((G*m*m_e)/F))/(G*m_e)
thats not it because i can't use m in the (U*sqrt((G*m*m_e)/F)) because m is not given in both equations they both involve m and r, and both m and r are not given in both equations, so i cant use what i posted in the third post

[Lemmy]

hehe i tried submitting the third post answer and its wrong :(

[Physics Monkey]

Lemmy, you have one equation there involving only m, you can solve it for m in terms of just G, m_e, F, and U.

Let me write it for you:

m = \frac{U \sqrt{G m_e m}}{\sqrt{F}}\frac{1}{G m_e},

can you see how to find m?

[Lemmy]

shouldn't the (U*sqrt((G*m_e*m)/F))) all be in the numerator?

[Physics Monkey]

Yes, sorry, I fixed it. Can you solve that for m? Hint: get all the m terms on one side.

[Lemmy]

http://img.photobucket.com/albums/v195/havokxz/equation1.jpg

i wrote it out in paint take a look

[Lemmy]

thing is physics monkey i can't change the left side term, because the problem has everything set to m= already and i can't change left, only answer i can submit is the right term

[Physics Monkey]

Yes you can, Lemmy. That's how solving equations works. Maybe what you end up with is \sqrt{m} equals something you know, so then you just take the square of both sides to find what m equals.

[Lemmy]

oh hmm i see, okay i'll try it, my algebra skills suck though

[Lemmy]

okay i solved and got m=(U^2)/(F*G*m_e)
can you check my algebra? i hope this is right *crosses fingers*

[Physics Monkey]

Looks fine to me.

[Lemmy]

okay going to give this a whirl thank you soo much for your help physics monkey, greatly appreciated

[Lemmy]

woooo got it right thanks physics monkey