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when a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as ? I am thinking of this problem with analogy to a hydrogen atom. For example when an electron is excited it gains energy and when it returns to the ground state it emits radiations. Or when current travels through a resistor its potential energy decreases as the positive charge carriers move through the resistor and this energy appears as heat in the resistor.

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  • $\begingroup$ What to you mean by "brought with uniform velocity"? Usually one would expect acceleration, so transformation from potential to kinetic energy. Eventually on impact everything will be transformed into heat. $\endgroup$ – mikuszefski Apr 19 '15 at 14:26
  • $\begingroup$ i mean if we apply force on the mass in the opposite direction of the gravitational field so that the mass doesn't accelerate, i.e we do work to transfer energy away from the system, where does this energy end up? $\endgroup$ – Musa Tibetbaqal Apr 19 '15 at 14:29
  • $\begingroup$ In whatever mechanism you use to keep it at constant speed. $\endgroup$ – mikuszefski Apr 19 '15 at 14:35
  • $\begingroup$ its just an external force on the particle, its source is not defined, can be anything. $\endgroup$ – Musa Tibetbaqal Apr 19 '15 at 14:43
  • $\begingroup$ Yes, can be anything, and this anything is taking the energy. $W=\int \mathrm{d}\vec x\cdot \vec F$. If it is e.g. a rocket it goes into the kinetic energy of the gas that the rocket ejects. $\endgroup$ – mikuszefski Apr 19 '15 at 15:04
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Ok. I see the confusion. See, when you bring a ball from infinity to a position at a distance r from the center of the earth with uniform velocity, you need to apply some external force on it ; otherwise naturally the ball would be accelerated (because of gravitational force of earth ) and the decreased potential energy would appear as the kinetic energy of your ball. But when you are applying some external force to keep the ball going with uniform velocity, you are actually transferring the energy from 'earth + ball' system to the external system applying external force on the ball. So the decreased potential energy appears as some form of energy - may be kinetic, potential, thermal or a mixture of all these forms of the external system used to apply force on the ball to keep it moving with uniform velocity.

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When a ball of mass m is brought with uniform velocity from infinity into the g field of the earth at a distance r from it, the potential energy of the ball earth system decreases from 0 to -GMm/r. What does this lost energy appear as?

Kinetic energy, which is typically radiated away into space. Imagine your ball starts off a long long way from Earth, and is initially motionless with respect to the Earth. Then it starts falling towards the Earth, slowly at first, then faster and faster. Eventually it's falling at 11.2 km/s. It hits the Earth with considerable kinetic energy, which is converted into heat and light etc. In an idealised scenario this is radiated away into space whilst the ball survives. Now the ball has the same velocity it used to have with respect to the Earth, namely zero. Bringing the ball to Earth with some uniform velocity also involves getting rid of kinetic energy, but gradually rather than all in one go at the end.

Note that conservation of energy applies, and that because the kinetic energy has been radiated away, you now have a mass deficit, see the Wikipedia binding energy article. The ball's mass is now less than m. Whilst conservation of momentum also applies, the Earth's motion is not discernible, and the kinetic energy and thence the mass deficit are not shared equally.

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I think what may be referred to in the question is the determination of gravitational potential energy (GPE). The definition for GPE is the work that was done in bringing the two masses to a distance r apart from an initial separation that was infinite. (Tsokos, 396) What is important here is that this happens at a very slow uniform speed so the Kinetic Energy does not affect the calculation.

The formula mentioned in the question assumes that an external agent provides the force to do the work on one of the masses. This force is equal and opposite to the force between the two masses. Essentially the mass is kept from free falling by delievering an opposite force. This is the reason why the equation has a negative sign infront of it. It shows that we needed to exert a force that was opposite. So the mass does not loose energy in that sense it simply requires energy to stop its free fall to determine accurately the potential.

Another much simpler explanation (which in essence describes the same thing) is that the negative sign represents the fact that Gravity is a force of attraction.

It is important that the situation is not comparable to the one mentioned with the hydrogen atom as they are governed by different mechanics.

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  • $\begingroup$ Yes. So the external agent results in the object having zero force, therefore no acceleration, no change in motion, and no change in energy. The agent however has to provide the $\int \mathrm{d} \vec x \cdot \vec F$, which is exactly the energy the object would have gained if in free fall. $\endgroup$ – mikuszefski Apr 19 '15 at 15:31
  • $\begingroup$ Exactly, the forces need to be held in equlibrium to allow for accurate calculations $\endgroup$ – Jaywalker Apr 19 '15 at 15:32
  • $\begingroup$ yes the total work done on the object is zero because the external force is stopping the object from gaining the energy which it would gain in free fall, so the external force takes away this energy. So this implies the mechanism which is being used to apply the external force ends up gaining the energy in some other form like thermal energy. Am i correct? $\endgroup$ – Musa Tibetbaqal Apr 20 '15 at 5:29

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