Do bodies lose mass due to gravity?

[Allen_Wolf]

Energy is required to make a body fall towards its surface i.e. to produce acceleration. That means energy will be lost by the body causing it. Energy is equivalent of mass. Therefore a loss in energy would mean a loss in mass. Right?
I most sincerely apologise of I am wasting your time by asking a possibly stupid question.
I just want to get things clear.☺

 

[thebosonbreaker]

Objects don't lose mass because they don't "lose energy".
Whilst you're correct in saying that energy is required to cause acceleration (work has to be done) and that mass and energy are linked (not so much "equivalent"), the energy is not "lost" due to the principle of the conservation of energy. Energy cannot be created nor destroyed but only transformed from one form to another.
The energy applied which is causing acceleration is transformed to kinetic energy as the body falls and possibly some other wasted forms such as heat or sound. Therefore the energy is conserved and thus the mass does not change.

 

[Allen_Wolf]

Objects don't lose mass because they don't "lose energy".
Whilst you're correct in saying that energy is required to cause acceleration (work has to be done) and that mass and energy are linked (not so much "equivalent"), the energy is not "lost" due to the principle of the conservation of energy. Energy cannot be created nor destroyed but only transformed from one form to another.
The energy applied which is causing acceleration is transformed to kinetic energy as the body falls and possibly some other wasted forms such as heat or sound. Therefore the energy is conserved and thus the mass does not change.

But there is a transfer of energy from , let's say the earth, when attracting a ball.Some of the energy of the Earth is tranferred to the ball and then transforms to kinetic energy. That means that the amount of energy possesses by the Earth now is less than the original value, right?

 

[thebosonbreaker]

But there is a transfer of energy from , let's say the earth, when attracting a ball.

I think "energy" may be the wrong term here. The ball will only fall towards the Earth due to gravity. The Earth of course possesses a gravitational field and so a force will act on the ball pulling it towards the Earth's centre.

Some of the energy of the Earth is tranferred to the ball

Whilst the ball will gain kinetic energy none of the "earth's energy" as such will be transferred, but rather, going back to the principle of conservation of energy I talked about earlier, the gravitational potential energy which the ball has due to being present in Earth's gravitational field will be converted to kinetic energy if the ball is not being stopped by another force.

That means that the amount of energy possesses by the Earth now is less than the original value, right?

Like I say, there is no net gain or loss of energy because this is impossible - so the energy posessed by the Earth will be unchanged.

That's about the best I can explain it, I hope this clears things up for you to at least some extent :)

 

[Allen_Wolf]

Thanks a lot. It was really helpful☺

 

[PeroK]

But there is a transfer of energy from , let's say the earth, when attracting a ball.Some of the energy of the Earth is tranferred to the ball and then transforms to kinetic energy. That means that the amount of energy possesses by the Earth now is less than the original value, right?

First, let's stick to classical physics. The Earth and the ball can be considered as a system If we have an initial state where the ball is held some distance above the Earth, then there is no Kinetic Energy (KE) in the system, but the ball has Gravitational Potential Energy (PE). For distances close to the Earth, this is simply $mgh$, where $m$ is the mass of the ball, $g$ is the Earth's gravitational acceleration at the surface and $h$ is the height above the surface.

If the ball is released, then the PE is transformed into KE of the ball. If we ignore air resistance, then all the PE goes to KE. And, if $v$ is the speed of the ball when it hits the surface we have:

$\frac12 mv^2 = mgh$

Which equates the KE of the ball with the PE it has lost by falling.

 

[Khashishi]

Energy is required to make a body fall towards its surface i.e. to produce acceleration. That means energy will be lost by the body causing it. Energy is equivalent of mass. Therefore a loss in energy would mean a loss in mass. Right?

In Newtonian physics, energy and mass are not equivalent, so there is no loss of mass. Falling converts potential to kinetic energy, but doesn't change the mass. But that's not the answer you are looking for, is it?

Things are more tricky when you consider the general theory of relativity (GR). Energy depends on your frame of reference. This is true in Newtonian physics as well as relativity. But in GR, a frame of reference only covers a very small region around a point, whereas in special relativity and Newtonian mechanics, it covers all space. Also, in GR, an inertial frame of reference is in free-fall. If you follow an object while it falls, there is no acceleration, so the kinetic energy is zero and the mass stays conserved. A frame of reference that is not in free-fall, such as an observer standing on the Earth, is not an inertial reference frame, so energy is not conserved.

Nevertheless, there are still particular quantities that are conserved which are closely related to energy and mass, so you can't use gravity to build a perpetual motion machine. You can extract energy from something falling into a black hole, but it will decrease the final mass of the black hole. The mass of a black hole is not equal to the mass of everything that fell into a black hole. When things fall into a black hole, they fall into an accretion disc where they are heated by collisions with other falling matter, and a lot of radiation is emitted. The energy of the radiation is a substantial fraction (4-49%) of the original mass of the infalling matter, so the black hole only gains mass equal to the remaining mass that wasn't emitted. We can define a mass for the black hole because it is more or less stationary to a faraway observer. But we can't give a unique definition for the mass or energy of the stuff as it falls in.

 

[jartsa]

But there is a transfer of energy from , let's say the earth, when attracting a ball.Some of the energy of the Earth is tranferred to the ball and then transforms to kinetic energy. That means that the amount of energy possesses by the Earth now is less than the original value, right?

Case1: A soft non-elastic ball falls on a hard ground. Ball gets warm.

Case2: A hard ball falls on the ground, the ground is soft and non-elastic. Ground gets warm.

We can be sure that the balls in the two cases have different masses, because they have different energies.
And we can be sure that the Earth's in the two cases have different masses, because they have different energies.

Now I leave as an exercise for the reader to decide what can be deduced from that.

 

[jartsa]

Case1: A soft non-elastic ball falls on a hard ground. Ball gets warm.

Case2: A hard ball falls on the ground, the ground is soft and non-elastic. Ground gets warm.

We can be sure that the balls in the two cases have different masses, because they have different energies.
And we can be sure that the Earth's in the two cases have different masses, because they have different energies.

Now I leave as an exercise for the reader to decide what can be deduced from that.

Well maybe nothing can deduced from that. I have not come up with anything anyway. But the following is actually a good point, I think:

If mass that ends up deeper in a gravity well loses energy, then:

1: ball that has moved closer to the ground has lost energy, as it is now deeper in the gravity field of the earth.
2: Earth that has been moved closer to by the ball has lost energy, as it is now deeper in the gravity field of the ball.

 

[jbriggs444]

If mass that ends up deeper in a gravity well loses energy, then:

Treating gravitational potential energy as if it were an attribute of an object is a trick that happens to work adequately on the Earth.

A more proper approach (at least classically) is to treat GPE as an attribute of a pair of objects and their shared gravitational attraction rather than of either object individually.

 

[Khashishi]

For weak gravity, you can treat the gravitational field as having a negative energy density analogous to the energy in the electromagnetic field. In which case, we treat the energy as existing in space and not an attribute of the objects. This is only a conceptual difference, not a physical one, since we don't have any way of distinguishing these pictures physically.