Angular momentum of a satellite

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1. May 8, 2016

RoboNerd

1. The problem statement, all variables and given/known data
A satellite is in a circular orbit of radius R from the planet's center of mass around a planet of mass M.

The angular momentum of the satellite in its orbit is:
I. directly proportional to R.
II. directly proportional to the square root of R
III. directly proportional to the square root of M.

The correct answer apparently is II and III.
2. Relevant equations

3. The attempt at a solution
OK, I thought that since the satellite's angular momentum is mvR * sin(90), then its angular momentum would be directly proportional to its mass, velocity, and/or even the radius orbit.

I was thus the only thing that made sense.

Why is the correct answer II and III? Thanks so much in advance for the help!

2. May 8, 2016

drvrm

angular momentum is moment of momentum , in a way you may have correctly thought of the dependence ;
but is there a dependence of v on R?
may be v is proportional to R^ -1/2 ! then the ang. momentum can be said to be dependent on R^1/2

Last edited: May 8, 2016
3. May 8, 2016

PeroK

This is a slightly nasty question because, in general, angular momentum is directly proportional to $r$ and $v$. So, if the question were:

An object is moving past a planet at a distance $R$, its angular momentum is ...? Then "proportional to R" would be correct.

But, in a circular orbit, $v$ depends on $R$ and the mass of the planet and the gravitational constant. So, it was necessary to express angular momentum in terms of all the known variables first.

The question is a bit tricky in my opinion. Slightly disengenuous, perhaps.

4. May 8, 2016

RoboNerd

If I have G * mass planet * mass sattelite / r^2 = mass sat. * v^2 /r

Then I have v = sqrt( G * mass planet/ r).

So the velocity is not directly proportional to the square root of r, but inversely proportional! Right?

5. May 8, 2016

Yes.

6. May 9, 2016

drvrm

your angular momentum is equal to m.v. r and if you say v to be proportional to 1/sqrt(r) ,

then you get the ang. momentum as directly proportional to sqrt(r) ;
i think that is what you need .
regarding dependence on mass - if you replace v in terms of mass and r which are under sqrt sign ---one can say that angular momentum is proportional to sqrt(mass) also

-

so at least it corroborates the answer

7. May 9, 2016

RoboNerd

No, I think it would be inversely proportional???

8. May 9, 2016

drvrm

so , again i repeat for your consideration:
L = m.v.r ,
v = constant / sqrt(r) ; substitute in L (ang. momentum)
L= m. constant / sqrt(r) . r = m . constant. sqrt(r) so L is proportional to sqrt(r)
as r = sqrt(r). sqrt(r)