# Escape velocity and kinetic energy of the Earth

• Alexander350
In summary: 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.
Alexander350
If you had two masses, $m_{1}$ and $m_{2}$, and you released them in space infinitely far apart, their kinetic energies would satisfy $\frac{1}{2}m_{1}v_{1}^2+\frac{1}{2}m_{2}v_{2}^2=\frac{Gm_{1}m_{2}}{r}$ 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?

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.

Alexander350
Alexander350 said:
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.

Alexander350
PeroK said:
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?

Alexander350 said:
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.

PeroK said:
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.

Alexander350 said:
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".

Alexander350

## 1. What is escape velocity?

Escape velocity is the minimum speed at which an object needs to travel to break free from the gravitational pull of a planet or other celestial body. For Earth, the escape velocity is about 11.2 kilometers per second.

## 2. How is escape velocity calculated?

The formula for calculating escape velocity is given by v = √((2GM)/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the object. For Earth, the mass (M) is 5.97 x 10^24 kilograms and the radius (r) is 6,371 kilometers.

## 3. What is the kinetic energy of Earth?

The kinetic energy of Earth is the energy an object possesses due to its motion. For Earth, this energy is constantly changing due to its rotation and orbit around the Sun. The total kinetic energy of Earth can be calculated by adding the kinetic energy of its rotation and orbital motion.

## 4. How is the kinetic energy of the Earth related to its escape velocity?

The kinetic energy of an object is directly proportional to its velocity squared. This means that as the velocity of an object increases, so does its kinetic energy. Since escape velocity is the minimum speed needed to break free from a planet's gravitational pull, it is directly related to the kinetic energy needed to overcome that gravitational force.

## 5. What are some real-world applications of understanding escape velocity and kinetic energy of the Earth?

Knowledge of escape velocity and kinetic energy of the Earth is important in space exploration and satellite launches. It helps scientists and engineers determine the necessary velocity and energy needed to launch spacecrafts into orbit and beyond. Understanding these concepts also plays a role in predicting the trajectories of objects in space and ensuring the safety of satellites and other space debris in Earth's orbit.

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