Elastic gravitational collision

In summary, the conversation discusses an exercise involving an elastic gravitational collision between a planet and a satellite. The question asks to show that the final velocity of the satellite is equal to twice the orbital speed of the planet. The hint is to translate the problem into the center of mass reference frame. There is a discussion about whether the change in momentum of the satellite can be ignored, as well as the role of the planet's "infinite" mass. The conversation ends with a summary of the equations for momentum and energy in an elastic collision and how they apply to this problem.
  • #1
GwtBc
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Member advised to use the formatting template for questions posted to the homework forums
Hi all.
Our lecturer gave us an exercise the other day regarding an elastic gravitational collision between a planet and a satellite where the satellite slingshots using the gravitational field of the planet. The question asks to show that ##v_{f} - v_{i} = 2v_{0}## where ##v_{f}## is the final velocity of the satellite, ##v_{i}## is it's initial velocity and v_{0} is the orbital speed of the planet. The hint is to translate this problem into the COM reference frame, which I did do, and got:

##p'_{i} = m(v_{i}- v_{0})##

and

##p'_{f} = m(v_f + v_0)##

. I was told by someone that I can't do this since the planet isn't an inertial frame, I thought it would be ok since the acceleration of M is extremely small,. But I'm still not sure since if the equations for the momentum of the system from the lab's reference frame are written out and then translated into the planet's reference then there's an ##M\delta v## term in the second equation which is absurd since the planet can't be moving in it's own ref frame. Also this implies that ##v_{f} + v_{0}## and ##v_{i} - v_{0}## are equal whilst they're in opposite directions. I hope someone can set me on the right track .
 
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  • #2
GwtBc said:
Hi all.
Our lecturer gave us an exercise the other day regarding an elastic gravitational collision between a planet and a satellite where the satellite slingshots using the gravitational field of the planet. The question asks to show that ##v_{f} - v_{i} = 2v_{0}## where ##v_{f}## is the final velocity of the satellite, ##v_{i}## is it's initial velocity and v_{0} is the orbital speed of the planet. The hint is to translate this problem into the COM reference frame, which I did do, and got:

##p'_{i} = m(v_{i}- v_{0})##

and

##p'_{f} = m(v_f + v_0)##

. I was told by someone that I can't do this since the planet isn't an inertial frame, I thought it would be ok since the acceleration of M is extremely small,. But I'm still not sure since if the equations for the momentum of the system from the lab's reference frame are written out and then translated into the planet's reference then there's an ##M\delta v## term in the second equation which is absurd since the planet can't be moving in it's own ref frame. Also this implies that ##v_{f} + v_{0}## and ##v_{i} - v_{0}## are equal whilst they're in opposite directions. I hope someone can set me on the right track .
You can consider the planet as of infinite mass. The energy is conserved in gravitational interaction , so the speed of the satellite with respect to the planet remains the same, while the velocity changes sign. Initially the satellite moves toward the planet, finally it goes away from it. See:https://en.wikipedia.org/wiki/Gravity_assist

200px-Gravitational_slingshot.svg.png
 
  • #3
Right. I had realized that the conservation of kinetic energy meant that the magnitude of the two velocities would be the same, but I just found it hard to convince myself that the change in momentum of the satellite can be ignored in the calculations, but if it has 'infinite' mass then that provides a sort of momentum sink right?
 
  • #4
GwtBc said:
Right. I had realized that the conservation of kinetic energy meant that the magnitude of the two velocities would be the same, but I just found it hard to convince myself that the change in momentum of the satellite can be ignored in the calculations, but if it has 'infinite' mass then that provides a sort of momentum sink right?
It is the change of velocity of the planet that is ignored because of its "infinite" mass.
 
  • #5
GwtBc said:
I just found it hard to convince myself that the change in momentum of the satellite can be ignored in the calculations, but if it has 'infinite' mass then that provides a sort of momentum sink right?
No. When one mass is very much larger than the other, we can ignore the KE change of the larger. If the mass ratio is M >> 1, momentum conservation says the the ratio of velocities is M (favouring the smaller mass) so the ratio of energies is also M, favouring the smaller mass.
The change in momentum of the planet was not ignored in ehild's analysis; indeed, it must be equal and opposite to the satellite's change in momentum.
 
  • #6
GwtBc said:
But I'm still not sure since if the equations for the momentum of the system from the lab's reference frame are written out and then translated into the planet's reference then there's an ##M\delta v## term in the second equation which is absurd since the planet can't be moving in it's own ref frame. Also this implies that ##v_{f} + v_{0}## and ##v_{i} - v_{0}## are equal whilst they're in opposite directions. I hope someone can set me on the right track .
To increase the speed of the satellite by slingshot, the satellite and the planet have to travel in opposite directions in the lab's frame of reference. If the orbital speed of the planet is U0 and the initial velocity of the satellite is vi along the positive x axis, the velocity of the planet is -U0.
In case of elastic collision between two bodies with masses m1 and m2, initial velocities V1i and V2i, the equations for momentum and energy are
m1V1i+m2V2i=m1V1f+m2V2f
m1V1i2+2V2i2+m1V1f2+2V2f2
Rearrange the equations to collect the '1' terms at one side and the '2' terms at the other side:
m1(V1i-V1f)=-m2(V2i-V2f) (1)
m1(V1i2-V1f2)=-m2(V2i2-V2f2)
Divide the second equation by the first one. You get
(V1i+V1f)=(V2i+V2f), (2) so
V1f=-V1i+(V2i+V2f).
In this problem, '1' means the satellite, and '2' means the planet. V2i=-u0 as they travel at opposite directions.
m1/m2<<1, you can see from equation (1) that V2i≈V2f = - u0.
The satellite travels with higher speed in the opposite direction after the slingshot.
 

1. What is an elastic gravitational collision?

An elastic gravitational collision is a type of collision between two objects in which both kinetic energy and momentum are conserved. This means that the total energy and total momentum of the system before and after the collision are the same.

2. How is elastic gravitational collision different from inelastic collision?

In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy such as heat or sound. In an elastic gravitational collision, all of the kinetic energy is conserved and there is no loss of energy.

3. What factors affect the outcome of an elastic gravitational collision?

The outcome of an elastic gravitational collision is affected by the masses and velocities of the objects involved. The angle of collision and the strength of the gravitational force between the objects also play a role.

4. How is the conservation of momentum and energy demonstrated in an elastic gravitational collision?

The conservation of momentum is demonstrated by the fact that the total momentum of the system before the collision is equal to the total momentum after the collision. The conservation of energy is demonstrated by the fact that the total energy of the system before the collision is equal to the total energy after the collision.

5. What are some real-life examples of elastic gravitational collisions?

Some examples of elastic gravitational collisions include billiard balls colliding on a pool table, two objects colliding in space due to gravitational forces, and the collision of particles in a particle accelerator. These collisions demonstrate the conservation of energy and momentum in different ways.

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