Escape velocity and kinetic energy of the Earth

  • #1
If you had two masses, [itex]m_{1}[/itex] and [itex]m_{2}[/itex], and you released them in space infinitely far apart, their kinetic energies would satisfy [itex]\frac{1}{2}m_{1}v_{1}^2+\frac{1}{2}m_{2}v_{2}^2=\frac{Gm_{1}m_{2}}{r}[/itex] if they met with a distance r between their centres of mass. This equation therefore tells you the velocities needed for the two bodies to escape the gravitational pull of each other, i.e. the escape velocities. So, why does the formula for the escape velocity of an object on earth only include the kinetic energy of the object, and not the earth itself? Is the kinetic energy of the earth just negligibly small (because the conservation of momentum means its velocity is pretty much zero) and can therefore be ignored? Would it only be necessary if the two objects had similar mass?
 

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  • #2
PeroK
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The equation you are used to for the gravitational potential assumes a small mass and a large mass, where the large mass is assumed not to move. If you have two masses of a similar relative size, then you have a different equation for the gravitational potential of the system. You might like to derive it as a useful exercise.
 
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sophiecentaur
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Is the kinetic energy of the earth just negligibly small (because the conservation of momentum means its velocity is pretty much zero) and can therefore be ignored?
The momentum is proportional to the velocity but the KE is proportional to the velocity squared. So the sharing of the available energy into the KE of each object is proportional to the square of the ratio of the velocities and inversely with the ratio of the masses. Velocity wins.
 
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  • #4
The equation you are used to for the gravitational potential assumes a small mass and a large mass, where the large mass is assumed not to move. If you have two masses of a similar relative size, then you have a different equation for the gravitational potential of the system. You might like to derive it as a useful exercise.
I have been thinking about this for a while and am still not sure how to derive it. The problem is that both the objects move different distances, so how can you calculate the work done on them?
 
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PeroK
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I have been thinking about this for a while and am still not sure how to derive it. The problem is that both the objects move different distances, so how can you calculate the work done on them?

Look at how the force acts over a small distance. Don't forget conservation of momentum, which relates the velocities of the two masses.
 
  • #6
Look at how the force acts over a small distance. Don't forget conservation of momentum, which relates the velocities of the two masses.
Even trying to use conservation of momentum, I always end up with the kinetic energy of one of the masses equalling the potential energy of the system. This cannot be right as then there would be twice as much energy as there actually is. How do I set up the equations as I am obviously approaching it wrong.
 
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PeroK
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Even trying to use conservation of momentum, I always end up with the kinetic energy of one of the masses equalling the potential energy of the system. This cannot be right as then there would be twice as much energy as there actually is. How do I set up the equations as I am obviously approaching it wrong.

If you want to work this out yourself you should post it as homework and show your working.

Otherwise you could look up "reduced mass".
 
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