# Conservation of energy given the Earth's rotation?

So earth and everything is spinning around at 1000mph at the equator. From our perspective we are at a standstill. But let's say a fighter jets flies east to West against earth at same speed, so now relative to someone in space, the earth spins but the plane is at a standstill.

Wouldn't that plane now lost its speed from the earth, going against conservation of energy? Conservation of energy always holds true, so what am I missing or is this a bad question.

Just curious. Thanks for any help. I've confused myself

gmax137
energy is a reference frame dependent quantity.

• vanhees71
jbriggs444
Homework Helper
So earth and everything is spinning around at 1000mph at the equator. From our perspective we are at a standstill. But let's say a fighter jets flies east to West against earth at same speed, so now relative to someone in space, the earth spins but the plane is at a standstill.

Wouldn't that plane now lost its speed from the earth, going against conservation of energy? Conservation of energy always holds true, so what am I missing or is this a bad question.
Conservation of energy holds true. It holds true regardless of what frame of reference you use to do the energy accounting in. But energy is not "invariant". How much energy an object has is not a property of the object. It is a property of the object in the context of a reference frame. How much an energy an object has depends on what standard of rest you use.

Let us simplify things a bit. Instead of a jet, we will imagine an automobile.

First adopt a frame of reference where the Earth is at rest and watch the car accelerate from a stop. The car accelerates forward. Momentum is conserved. The Earth accelerates rearward. The Earth does not move very rapidly rearward since it is so enormously more massive than the car.

The car gains energy due to its forward motion. If we did the calculation, we would see that the gain in the Earth's energy due to its rearward acceleration is negligible.

Now step back and adopt a frame of reference where the Earth is moving rearward and watch the car accelerate from rest relative to the Earth. The car accelerates forward. Momentum is conserved. The Earth accelerates rearward. The car has lost kinetic energy. But the Earth has gained kinetic energy. If we did the calculation, we would see that the loss in the Earth's energy due to its rearward acceleration is NOT negligible.

In fact, the net gain in kinetic energy for the Earth+car system is identical no matter what frame of reference we use when performing the calculation.

Do the calculation yourself. Let the mass of the earth be M, the mass of the car be m and the forward velocity gain by the car be v. Compute the change in energy for three scenarios.

1. Car and Earth start with zero velocity. Final energy gain = ???
2. Car and Earth start with velocity -100v. Final energy gain = ???
3. Car and Earth start with velocity +100v. Final energy gain = ???

• vanhees71
So Jbriggs44 would it be safe to say my question is kinda poor as I'm confusing the calculation by changing reference frames? From what your saying that the issue, I nee to look at both reference frames and not mix the two correct?

Also I think I understand what you mean by invariant, but could you give me your definition so I know for sure? Thank you

jbriggs444
Homework Helper
So Jbriggs44 would it be safe to say my question is kinda poor as I'm confusing the calculation by changing reference frames? From what your saying that the issue, I nee to look at both reference frames and not mix the two correct?
You only need one reference frame. Any one will do. But using a quantity from one frame in a calculation done in another is a no-no.

I think that your original concern could be stated: "The plane's kinetic energy increased in this frame, therefore its kinetic energy increased in this other frame but actual calculation shows that it decreased instead -- what's wrong?". Obviously you were right to be concerned about the correctness of your reasoning.

Also I think I understand what you mean by invariant, but could you give me your definition so I know for sure? Thank you
A quantity is "invariant" if it has the same value no matter what frame of reference you use to measure it.

This is distinct from "conserved". A quantity is "conserved" if it has the same value no matter when you measure it [from a particular frame of reference].

If you perform the calculations that I suggested, you may be able to see an example of an invariant quantity.

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• vanhees71 and alkaspeltzar
Jbrigg 44, I guess what still confuses me is where did energy go?

Let's say a bus moves forward with a constant velocity and I walk backwards within the bus at a constant velocity. From the outside ground perspective, it looks like I'm moving slower and thus my original kinetic energy is gone.

so how does energy balance work here. I get it from inside the bus. I burn chemical energy to speed up relative to bus. But from outside that energy appears to be lost?

jbriggs444
Homework Helper
You have not done the exercise that I set.

vanhees71
Gold Member
You have to analyse the situation you describe cleanly in one and only one reference frame at any time. So just do the energy balance equation first in the rest frame of the bus and then in the rest frame of the street. There's no contradiction there, because there's a one-to-one mapping between the physical quantities as measured in the one or the other frame of reference. The natural laws are the same described in both frames of reference (here it's the symmetry of the laws of Newtonian mechanics under Galileo transformations).

This is an utmost important topic. The entire success of modern physics is due to the discovery of the importance of symmetry principles by Einstein et al at the beginning of the 20th century, leading to one of the most beautiful results of mathematical physics by Noether in discovering the close relation between symmetries and conservation laws.

So I'd also suggest to take @jbriggs444 advice seriously and do the calculation carefully in both frames of reference!

You have not done the exercise that I set.
I guess it's becuase I just don't understand what numbers I'm plugging in and where based on each reference frame. Like the -100 v, does that mean both car and earth are at -100. So if I accelerate to 100 v, do I use 100? Sorry I'm not that advanced.

I guess it's becuase I just don't understand what numbers I'm plugging in and where based on each reference frame. Like the -100 v, does that mean both car and earth are at -100. So if I accelerate to 100 v, do I use 100? Sorry I'm not that advanced.
Do I just use 1/2mv2...sorry been 15 years since I took physics in college. Hoping someone could break it out for me. I don't remember the math that well

vanhees71
Gold Member
Another good advice is to first write down everything in symbols and only at the end plug in the quantities (carefully using the correct units as well!).

jbriggs444
Homework Helper
I guess it's becuase I just don't understand what numbers I'm plugging in and where based on each reference frame. Like the -100 v, does that mean both car and earth are at -100. So if I accelerate to 100 v, do I use 100? Sorry I'm not that advanced.
Fair enough. Let's start with the easy one. Mass M earth at speed 0. Mass m you at speed 0. By inspection, initial energy is zero.

You push off from the earth gaining a speed of +v relative to your starting condition. The earth recoils at some speed. Calculate the resulting speed at which it recoils. Now calculate the final kinetic energy of earth and of you. Add them together.

Subtract initial total kinetic energy from final kinetic energy. What was the delta?

Now repeat. But the starting condition is you and Earth at -100v. After you push off, you are at -99v and the Earth is at some greater rearward speed. Calculate the total final kinetic energy. Calculate the difference between total final kinetic energy and total initial kinetic energy. What was the delta?

Repeat again. But now the starting condition is you and Earth at +100v. After you push off, you are at +101v and the Earth is somewhat less. What was the delta this time?

@vanhees71 would have us make the initial shared velocity another symbolic quantity. It is tempting to set the problem that way, but some people feel more comfortable with numbers than symbols. Possibly do the exercise a fourth time, but let upper case V stand in for the shared starting velocity.

• vanhees71
Let's look at example where both are at +100v and car pushes off to be at 101v. Why would earth speed change. Wouldn't I still see the earth at 100v?

Or again if both car and earth are at -100v, if car speed up to be at -99v, wouldn't I see earth at 100v? When I do math I see a loss of 1/2 mv2 using 1v as the difference. I'm seeing something wrong

jbriggs444
Homework Helper
Let's look at example where both are at +100v and car pushes off to be at 101v. Why would earth speed change. Wouldn't I still see the earth at 100v?
Conservation of momentum. If you move forward, the Earth has to move rearward. I gave you symbols to use for your mass and the mass of the Earth. Can you use them to compute the recoil speed of the Earth?

jbriggs444
Homework Helper
When I do math I see a loss of 1/2 mv2 using 1v as the difference. I'm seeing something wrong
Note that the difference between ##\frac{1}{2}m(-100v)^2## and ##\frac{1}{2}m(-99v)^2## is not ##\frac{1}{2}mv^2##.

I think I'm seeing that there is alot more too it than just 1/2mv2 and calculating the energy from simple speeds. I think I'm over simplifying it and without the understanding you folks have it is confusing me, hence why I see contradictions. I would probably have to have it pictorially drawn out to see it as it is kinda over my head..

In general though, I think I see what I am missing. Regardless of how you view a system, energy is conserved. But depending on the system and reference frame, you have to take other things/motions into perspective. Without doing so, looks like I'm missing energy. It isnt, just elsewhere in the problem.

With that I'm just going to fall back to basics and my understanding of conservation of energy. I appreciate the help. Thank you

jbriggs444
Homework Helper
I think I'm seeing that there is alot more too it than just 1/2mv2 and calculating the energy from simple speeds. I think I'm over simplifying it and without the understanding you folks have it is confusing me, hence why I see contradictions. I would probably have to have it pictorially drawn out to see it as it is kinda over my head..

In general though, I think I see what I am missing. Regardless of how you view a system, energy is conserved. But depending on the system and reference frame, you have to take other things/motions into perspective. Without doing so, looks like I'm missing energy. It isnt, just elsewhere in the problem.

With that I'm just going to fall back to basics and my understanding of conservation of energy. I appreciate the help. Thank you
You are correct. In a nutshell, you have to account for the energy going into the recoil of the Earth. Or the bus. Or the rocket ship exhaust. The recoil energy will vary depending on what reference frame you use for the calculation.

• alkaspeltzar
I guess that's all I want to know. Is energy always conserved? My engineering background says yes. When I change reference frames it looks different. I assume its still there, just how you relate/view it

jbriggs444
Homework Helper
I guess that's all I want to know. Is energy always conserved? My engineering background says yes. When I change reference frames it looks different. I assume its still there, just how you relate/view it
Yes. Once you pick a reference frame, energy is conserved in that frame.

• alkaspeltzar
You are correct. In a nutshell, you have to account for the energy going into the recoil of the Earth. Or the bus. Or the rocket ship exhaust. The recoil energy will vary depending on what reference frame you use for the calculation.
They never gave us examples this deep in school. Our mechanics and energy balance examples were very simple. As is the hydraulics I work on. Guess they didn't want to confuse us, rather take their word conservation of energy is true, haha don't worry about the rest.

jbriggs444