• Support PF! Buy your school textbooks, materials and every day products Here!

Angular momentum of a satellite

  • #1
410
11

Homework Statement


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.

Homework Equations




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!
 

Answers and Replies

  • #2
963
213
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!
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:
  • #3
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
12,302
5,111

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!
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
410
11
may be v is proportional to R^ -1/2 ! then the ang. momentum can be said to be dependent on R^1/2
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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,719
5,030
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?
Yes.
 
  • #6
963
213
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?
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

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.
-

so at least it corroborates the answer
 
  • #7
410
11
then you get the ang. momentum as directly proportional to sqrt(r) ;
No, I think it would be inversely proportional???
 
  • #8
963
213
No, I think it would be inversely proportional???
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)
 

Related Threads on Angular momentum of a satellite

Replies
16
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
9K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
7K
  • Last Post
Replies
3
Views
9K
Replies
5
Views
1K
Replies
2
Views
4K
  • Last Post
Replies
10
Views
5K
Top