Satellite velocity and momentum

In summary: The total energy immediately after the explosion is: .5m(vi)^2 - GM(4m)/2aThe length of the new orbit is: r*vi
  • #1
jlayla99
2
0

Homework Statement


A satellite moves around the Earth in a circular orbit of radius r.
a) What is the speed vo of the satellite?
Suddenly an explosion breaks the satellite into two pieces, with masses m and 4m. Immediately after the explosion the smaller piece of mass m is stationary with respect to the Earth and falls directly toward the Earth.
b)What is the speed vi of the larger piece immediately after the explosion?
c) Because of the increase in its speed this larger piece now moves in a new elliptical orbit. Find its distance away from the center of the Earth when it reaches the other end of the ellipse.


Homework Equations


K=.5mv^2
E=-GMm/2r
mv=mv
E=-GMm/2a



The Attempt at a Solution


a) I think I got the answer to a:
.5mv^2=GMm/2r
v=sqrt(GM/r)
Then substituting in the values of G and M (Mass of the Earth):
v=1.99x10^7/sqrt(r)

b) I am having trouble with this because I don't know if I can simply use linear momentum. If so then I can just assume that the smaller mass has no momentum in relation to the first.
mv=mv
5m(vo)=4m(vi)
(5/4)vo=vi
Is it really that simple?

c) I'm kind of shooting in the dark with this one, but can I assume that the energy initially (GM(5m)/2r) is going to equal the second energy (GM(4m)/2a?
 
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  • #2
jlayla99 said:
a) I think I got the answer to a:
.5mv^2=GMm/2r
v=sqrt(GM/r)
Then substituting in the values of G and M (Mass of the Earth):
v=1.99x10^7/sqrt(r)
OK, but I wouldn't bother plugging in numerical values. Leave your answer in terms of G, M, and r.

b) I am having trouble with this because I don't know if I can simply use linear momentum. If so then I can just assume that the smaller mass has no momentum in relation to the first.
mv=mv
5m(vo)=4m(vi)
(5/4)vo=vi
Is it really that simple?
Yes, that simple.

c) I'm kind of shooting in the dark with this one, but can I assume that the energy initially (GM(5m)/2r) is going to equal the second energy (GM(4m)/2a?
Compare the energy of the masses before and after the explosion.
 
  • #3
what do you mean compare the energies?
the energy before: .5m(vo)^2 - GM(5m)/2r
the energy after: .5m(vi)^2 - GM(4m)/2a

where do i go from there?
 
  • #4
Express the total energy immediately after the explosion in terms of the energy before the collision. Then use that new energy to solve for length of the new orbit.
 

FAQ: Satellite velocity and momentum

What is satellite velocity and why is it important?

Satellite velocity refers to the speed at which a satellite is moving in its orbit around a planet or object. It is important because it determines the satellite's orbital period, or the time it takes for the satellite to complete one full orbit. It also affects the satellite's ability to maintain its position in orbit and avoid collisions with other objects.

How is satellite velocity calculated?

Satellite velocity can be calculated using the formula v = √(GM/r), where v is the satellite's velocity, G is the universal gravitational constant, M is the mass of the planet or object the satellite is orbiting, and r is the distance between the satellite and the center of the planet or object.

How does satellite velocity relate to momentum?

According to Newton's second law of motion, momentum is equal to the mass of an object multiplied by its velocity. Therefore, the velocity of a satellite directly affects its momentum. The higher the velocity, the greater the satellite's momentum will be.

Can satellite velocity be changed?

Yes, the velocity of a satellite can be changed by applying a force to it. For example, a rocket can be used to boost a satellite's velocity and change its orbit. This is often necessary in order to maintain the satellite's desired position or to avoid collisions with other objects in space.

What factors affect satellite velocity and momentum?

The main factors that affect satellite velocity and momentum are the mass of the planet or object being orbited and the distance between the satellite and the center of that object. Other factors may include atmospheric drag, gravitational pulls from other nearby objects, and changes in the satellite's trajectory due to external forces.

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