# Work, Gravitational Potential and Kinetic Energy, Spacecraft

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1. May 1, 2017

### ashyyamora

• Does not use the template because it was originally posted in a non-homework forum
Hi guys. I'm new to this forum so I'm not sure if I'm posting this in the right place or correctly, but I was totally stumped on a Physics 12 question as I was studying for my exam.

The question is:
An explorer spacecraft is descending towards Mars using a rocket engine for braking. The spacecraft has Ep = -2.8x10^10 J and Ek = 3.0x10^9 J at its initial position. Some time later the spacecraft has Ep = -3.4x10^10 J and Ek =1.0x10^9 J. How much work did the rocket engine do in braking the spacecraft?

At first, I thought it was work done = change in kinetic energy, but that's not the answer. Then I realised that the change in kinetic energy and the change in potential energy were not the same, and that's where I was completely lost. What does it mean when its gravitational potential energy increases in the negative? Is it moving closer or farther from the planet? And how would I go about solving this question?

I tried (although I don't know why): Wbrakes - [(-3.4x10^10) - (-2.8x10^10)] = -2.0x10^9; Wbrakes = -2.0x10^9 - 6.0x10^9 = -8.0x10^9 J which is the correct answer. I just don't know if that's the correct solution and if it is, why.

2. May 1, 2017

Without any breaking, the rocket's total energy (sum of potential +kinetic) would remain the same. With braking applied, as the rocket approaches the planet and its kinetic energy becomes more negative, normally the change is all converted to kinetic energy to keep the total energy constant. Braking keeps this from occurring, and with sufficient braking, you could even get the rocket to land on the planet rather than crashing into it at high speed.

3. May 1, 2017

### ashyyamora

If the potential energy's magnitude is increasing, does it mean the spacecraft is getting closer or farther from Mars? Why does the braking affect the potential energy?

4. May 1, 2017

A more negative potential energy (larger in magnitude) means it is getting closer to the planet. Potential energy changes with distance $r$ from the center of the planet) by the formula $E_p=U=-GMm/r$ where $G$ is the gravitational constant $M$ is the mass of the planet and $m$ is the mass of the spacecraft. You can see as $r$ gets smaller that $U$ becomes more negative (larger in magnitude). $\\$ Editing... Meanwhile the braking (retro-rocket) affects the kinetic energy and thereby the total energy. The braking does not affect the potential energy=that is simply a function of the location ($r$). Again, without braking the total energy stays the same so that if the potential energy becomes large and negative, that means without braking that the kinetic energy will become large and positive. The braking lowers the energy and lowers the total energy as well as the kinetic energy. Of course if you fired the same (retro-)rocket in the other direction, you could actually increase the total energy and increase the kinetic energy tremendously.

Last edited: May 1, 2017
5. May 1, 2017

### ashyyamora

Okay I think I get it. To recap simply: the approaching of the spacecraft to the planet is increasing the magnitude of the potential energy, but it's negatively, meaning that the kinetic energy is increasing positively. So the braking is not only negating that gain in kinetic energy, but it's also making it lose an additional 2.0x10^9J of kinetic energy. So in total, the braking is negating 2.0x10^9J + the 8.0x10^9J of kinetic energy gained from approaching the planet (gained due to acceleration from gravity).

Does that sound right?

6. May 2, 2017

The 8.0 you mentioned should be a 6.0 ,$(3.4-2.8) \cdot 10^{10}=6.0 \,\cdot 10^9$, so that $2.0 \,\cdot 10^9 J+6.0 \cdot \,10^9 J=8.0 \cdot \, 10^9 J$.