My professor said that the potential energy of an object is larger the higher off the ground it is when it falls. I asked if there is a limit to that potential energy and she said no. I then asked about terminal velocity. Surely if two completely equal objects are dropped from heights sufficiently high enough for both of them to have reached terminal velocity, the height no longer matters pertaining to the amount of energy released? Whether one was up a mile further than the other, at some point they are both falling at the same rate and thus their potential energy is equal from that point on? In my mind, I then equate it to two equal objects falling at the same speed. They will both impact the ground with the same force, no? Where am I going wrong with this?

Edit: I appreciate the answers, but being a neophyte, I’m still slightly confused about the ultimate answer. My new understanding given the answers provided, is that the energy released by the object that has spent a longer time falling will be greater because it has had more energy gathered in some way due to having more time to heat up the air around it, and thus will actually impact with a greater energy released than the object with less time to fall. Is that correct?


Since you are talking about terminal velocity, you are including air resistance in this thought experiment. This means that there is another place for the potential energy of object to go. When an object reaches terminal velocity, it means that every bit of work done by gravity ($mg\Delta h$), goes into heating up the object and the air around it instead of increasing the velocity of the object. So, an object that falls from a greater height will reach the same terminal velocity as one that falls from a lesser height, but it will be hotter and have a greater amount of hotter air above it.

More technically, at terminal velocity, the decreasing potential energy of the object is accompanied by an increase in internal (thermal) energy of the object and air, so energy is conserved. The increase in internal energy is evidenced by the increased temperature of the object and the air it passes through

  • $\begingroup$ Heat is energy transfer due solely to temperature difference. Heat is not the energy itself but a mechanism for transferring energy, the other being work. Heat is not "conserved". The energy being transferred is conserved. Instead of "heating up", it's better to say the friction of air resistance raises the temperature of the object (its internal energy). Once the temperature of the object is raised above the temperature of its surroundings, then heat transfer can occur between the object and its surroundings. $\endgroup$ – Bob D Feb 4 at 15:10
  • $\begingroup$ @BobD Subtle point. I've edited my answer. $\endgroup$ – Mark H Feb 4 at 19:36
  • $\begingroup$ I like the last paragraph. But I still don’t like statement in the first paragraph that gravity is “heating up” the object. There is no transfer of energy to the object from the air or gravity due to a temperature difference. But that’s OK I don’t mean to belabor the point since the last paragraph does the job $\endgroup$ – Bob D Feb 4 at 21:15
  • $\begingroup$ @BobD Thanks for the comments. Thermodynamics was always my worst subject, so I appreciate corrections. $\endgroup$ – Mark H Feb 4 at 21:32
  • $\begingroup$ This is one of those cases where newtonian-mechanics and thermodynamics concepts intersect. Within mechanics it is common to talk about "friction heating", whereas in thermodynamics the term "friction heating" is frowned upon. Work and heat are the two mechanisms of energy transfer. When friction increases temperature (like when you vigorously rub your hands together to feel "warm") it is really energy transfer by friction work. Heat (and heating) is not involved. Similarly when a gas is compressed its temperature increases because of work. Compression does not "heat" the gas up. $\endgroup$ – Bob D Feb 4 at 22:27

@MarkH already answared, but I want to add, that there indeed is a limit to potential energy, but for different reasons than you are suggesting.

For small heights the formula for gravitational potential energy is given by $$E_p=mgh,$$ where $m$ is mass of the object, $g$ is gravitational acceleration and $h$ is the height from the ground. This formula is of course unbounded, but it works only for small values of $h$. For larger values, you can no longer consider $g$ to be constant and you must take its dependence on $h$ into an account (the further you are from Earth, the weaker the gravity is). The correct formula for $h$ that varies a lot is: $$E_p=-G\frac{Mm}{h+r_e},$$ where $G$ is gravitational constant, $M$ mass of the Earth and $r_e$ is Earths radius.

For $h$ small compared to $r_e$ you get from this formula: $$E_p\approx-G\frac{Mm}{r_e}\left(1-\frac{h}{r_e}\right)=\text{const}+mgh,$$ where $g=G\frac{M}{r^2_e},$ which is the reason why for small heights the first formula is used. The constant term is irrelevant, because for computations only differences in energy matter. But it is negative, so the potential energy is negative number and as you go higher and higher this negative number gets smaller and smaller. In the limit of infinite height, it becomes zero, so the potential energy is actually bounded.

P.S. The approximation is given by Taylor expansion, which tells you that for $x \ll 1$ it holds $(1+x)^{-1}\approx 1-x.$ You can try it on calculator for small values of x.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.