Potential energy and kinetic energy of a revolving object like moon. Whats there role

nishantve1

So lets consider the moon its rotating around the earth in a fixed orbit, its moving at a velocity say v so it possess a kinetic energy 1/2 mv² . the gravitational force between the earth and the moon is also present which attracts the moon towards the earth . My question is does the moon has a gravitational potential energy of mg where m is the mass of the moon and g is the value of gravitational acceleration in the space ? What role does this potential energy play, the kinetic energy keeps the moon moving and the centripetal force mv²/r keeps it in its orbit . So where does potential energy blends in ? Is this energy responsible for keeping the moon bounded to the earth ? If not then what energy keeps the moon bounded to earth ?
Thanks in advance

mfb

No, that formula works for small height differences only, where the variation of g is negligible (e.g. in your lab).

For satellites like the moon, the gravitational potential energy is given by ##E=- \frac{MmG}{r}## where M is the mass of earth, m is the mass of moon, r is the distance and G is the gravitational constant.
It is negative, indicating that the moon is attracted to earth. Moon is bound because the sum of potential energy and kinetic energy is negative (more specific: the kinetic energy is half the (negative) potential energy): You would need additional energy to remove the moon from earth.

nishantve1

So if I use energy equal to the potential energy , will I be able to remove the moon?(Sounds like some evil plan LOL)also if I consider potential energy to be 0 how would it affect the energy required to remove the moon?

mfb

You just need 50% of the potential energy to remove moon, as the other 50% are already there (as kinetic energy).

The moon does not care about your choice of potential energy. It is convenient to set "potential energy at infinite distance" to 0, but you can use every other value, too.

nishantve1

Cannot understand this , would mind explaining it once?

mfb

Gravitational binding energy of the moon: GMm/r = -8*10²⁸J
Kinetic energy of the moon: 1/2mv^2 = 4*10²⁸J (rough approximations)

Total energy of the moon: -4*10²⁸J

Minimal energy of the moon at "infinite" distance: 0

Required energy to remove moon: 0 - (-4*10²⁸J) = 4*10²⁸J
(This is about 10⁸ times the world energy consumption of a year)

The actual value is a bit smaller than that, as I did not take the sun into account - you don't have to move it to "infinite" distance, something like ~1.5 million km would be enough to separate it.