Kinetic to potential energy ratio for orbiting satellite

[Princess19]

Homework Statement

What is the ratio of kinetic energy to potential energy, of a satellite orbiting a mass M around a radius R.

Homework Equations

ke = 1/2 (mv^2)/r, pe = -Gmm/r^2 (im not really sure of these, i don't have my book with me)

The Attempt at a Solution

this was a problem on my test, and i don't know if i got it right, so i would like to know what the answer is.

i had no idea how to do this so i just did

2ke = pe

ke/pe = 1/2.

what is the answer? i know i got it wrong hehe

any input is appreciated! thanks!

 

[kuruman]

Start from
KE = (1/2)mv²
PE = -GMm/r

Write Newton's 2nd Law for a circular orbit and solve for the speed v.
Replace v in the KE equation.
Take the ratio and see what you get.

 

[Princess19]

all i want to know is if i have it right. were not going to have a final exam so i won't see this stuff ever again.

anyways ill try it because I am bored...

(1/2)mv2/-GMm/r

thats its i have no idea what else

 

[kuruman]

all i want to know is if i have it right. were not going to have a final exam so i won't see this stuff ever again.

You don't have it right.

 

[Princess19]

haha i knew it

so how do u do this problem? all i can get is to that first part i just did

 

[kuruman]

In my posting #2 I gave you three steps. You skipped to the third without doing the first two. OK, once more

Step 1
Write Newton's 2nd Law for a circular orbit and solve for the speed v.

Finish this and I will help you with the rest.

 

[Princess19]

i still don't know how to do it

 

[Princess19]

can someone help me?

 

[zachzach]

Maybe you should check your sign?

 

[Princess19]

im so lost, wat sign?

im up to this part

(1/2)mv^2/GMm/r

 

[kuruman]

i still don't know how to do it

Newton's 2nd Law says F = ma. For a satellite of mass m orbiting the Earth in a circular orbit, what is F and what is a?

 

[Princess19]

Newton's 2nd Law says F = ma. For a satellite of mass m orbiting the Earth in a circular orbit, what is F and what is a?

F = m.v^2 / r ??

 

[kuruman]

Correct for the ma part. Now what is the force F for the gravitational attraction?

 

[Princess19]

Correct for the ma part. Now what is the force F for the gravitational attraction?

mg ?

 

[kuruman]

No. It is the Law of Universal gravitational force that has upper case G not lower case g in it.

 

[Princess19]

F = GMm/r^2 ?

 

[kuruman]

That's the one. Now let's go to step 2. You have F (posting # 16) and you have ma (posting #12). Set the two equal and solve for the quantity mv².

 

[Princess19]

That's the one. Now lets' go to step 2. You have F (posting # 16) and you have ma (posting #12). Set the two equal and solve for the quantity mv².

ok so

F = ma
GMm/r^2 = m . v^2 /r

GMm/r = mv^2

is that it? what next?

 

[kuruman]

Step 3. You know that KE = (1/2)mv². What is KE in terms of GMm/r?

 

[Princess19]

Step 3. You know that KE = (1/2)mv². What is KE in terms of GMm/r?

i don't understand lol

is it GMm/2r?

 

[kuruman]

Right. KE = GMm/(2r). You know what PE (see posting #2).
So what is the ratio KE/PE?

 

[Princess19]

1/2? or is it 2?

 

[Princess19]

-1/2?

 

[kuruman]

Bingo.

 

[Princess19]

hmm weird that wasnt one of the choices on the test, just 1/2 and that's what i marked. wonder if its a typo on the test...
oh well thanks a bunch!

 

[kuruman]

Strictly speaking one of the choices should have been -1/2. If you marked 1/2 and that was the only choice, then you probably got full credit. Some examiners don't pay attention to signs, but that is confusing to students who do.