# Gravatational Force between earth and moon

1. Mar 22, 2009

### thomas49th

I was wondering, if the moon losses its mass at a constant rate and the earth gains mass at this constant rate (so the sum of there masses is equal), the force between them would stay the same would it not?

F = GMm/d²

What happens to the moon?

Does the moon move away, stay the same or come closer to the earth. Also would the effect of the moon losing mass mean that the tides of the sea seize to exist?

Thanks :)

Last edited: Mar 22, 2009
2. Mar 22, 2009

### Gear300

I haven't done the math, but the better way to say it is the product of the masses is equal. Classically viewing it, the centripetal acceleration of the moon is dependent on the earth's mass, in which as the earth gets bigger, the needed acceleration would increase. The moon's orbit would get smaller and closer to the earth.

3. Mar 22, 2009

### mathman

The force obviously changes as the Newton equation shows. The moon probably comes closer. The tidal effect is complicated, since the loss of moon's mass and the decrease in d have opposite effects.

4. Mar 23, 2009

### thomas49th

Would the product of the masses decrease (while the sum stays the same). Therefore the force decreases thus the distance increases? Therefore the moon moves out of orbit?

Not sure that is right, but could you show me your reasoning for the moon to move closer. Is that because the earth exerts a greater force on the moon.

Thanks :)

5. Mar 23, 2009

### Janus

Staff Emeritus
The force does decrease, as the product of the masses decreases as mass is moved from the smaller mass to the larger mass. The distance also decreases, as the speed needed fro orbit a a given distance (taking the mass of both bodies into account is found by:

$$V_o = \sqrt{\frac{GM^2}{r(M+m)}}$$

where M would be the mass of the Earth while m is the mass of the Moon. Note that while M+m remains constant, M increases as mass moves from M to m. Since we have to assume that the speed of m does not change, it would be less than the new speed needed to maintain that orbit and it will fall into a lower one.

6. Apr 7, 2009

### thomas49th

Hi, I'm not sure I fully understand it:

"Since we have to assume that the speed of m does not change, it would be less than the new speed needed to maintain that orbit and it will fall into a lower one"

are you saying the orbit speed of m around M is the same?

also when you said:

"M increases as mass moves from M to m"

do you mean when the mass moves from m to M

7. Apr 7, 2009

### Janus

Staff Emeritus
Assume that we are talking about the Moon at a certain point of its orbit, and it has a velocity of 1 km/sec. Now we move a certain percentage of the Moon's mass to the Earth. We assume that at this instant we have not changed the velocity of the Moon nor its distance from the Earth(r) by removing the mass, and that right afterward, it is still moving at 1 km/sec.

The Earth, however, has gained mass, so according the equation in my last post, the orbital velocity needed to maintain an orbit at a distance of r goes up to a little bit more than 1 km/sec. Since the Moon is still moving at its original velocity, it will be moving slower than it needs to hold its original orbit; it will begin to fall in towards the Earth. As it does so, it will gain speed, as it trades potential energy for kinetic energy.

At a certain point (half an orbit from where it started its fall inward), it will have picked up enough speed to be moving faster than orbital speed for its new, closer distance. It will begin to climb back out again, until it reaches the point in its orbit where it first started to fall inward. It then repeats the whole process over again. IOW, it will enter an elliptical orbit, one which has a smaller average orbital distance from the Earth than it had before we moved some of its mass to the Earth.
Yes that was a typo/