# Conservation of Energy with Gravitational Potential Energy

• zachary570
In summary, the author was questioning the methods that they had used in the past when solving problems. They realized that the total energy of the system remains constant and that they use the conservation of momentum to relate the two velocities. They also understand that the gravitation potential energy refers to the system of the two objects. However, when they attempted to solve the problem, they didn't realize that kinetic energy of one of the objects needs to be included. They were also confused about when it is appropriate to leave out the kinetic energy of one of the objects. The author provides a way to solve the problem without needing to ask that question. They also understand that the small mass has more kinetic energy than the Earth so the velocity of the small mass
zachary570
Homework Statement
An experiment is performed in deep space with two uniform spheres, one with mass ##m_{A}## and the other with mass ##m_{B}##. They have equal radii, ##r##. The spheres are released from rest with their centers a distance ##r_{i}## apart. They accelerate toward each other and. you may ignore all gravitational forces other than that between the two spheres. When their centers are a distance ##r_{f}## apart what is the speed of each sphere?
Relevant Equations
Conservation of Energy: ##K_{i} + U_{i} = K_{f} + U_{f}##
Conservation of Momentum: ##m_{A}v_{A} = m_{B}v_{B}##
Gravitational Potential Energy: ##U = \frac{-Gm_{A}m_{B}}{r}##
I was working on this problem but after getting to the answer I questioned the methods that I used for previous problems that I had solved. I understand that the total energy of the system remains constant and that we use the conservation of momentum to relate the two velocities. This gives two equations with two unknowns and the math is straight forward. However, when I first attempted this I didn't realize that you need to include the kinetic energy of both objects and went about solving it the usual way where that mass cancels and I got $$v_{A} = \sqrt{2Gm_{B}\left( \frac{1}{r_{f}} - \frac{1}{r_{i}} \right)}$$ I understand now that the gravitation potential energy refers to the system of the two objects as it has the mass of both objects in the equation. What I am confused about is when it is appropriate to leave out the kinetic energy of one of the objects. Is it just when one of the objects has a much larger mass than the other object so that the acceleration the larger mass has is so small that it is negligible? What would be best when working through a problem like this on a test and I need to show my work. Should I include the larger mass's kinetic energy when expanding ##K_{i} + U_{i} = K_{f} + U_{f}## but then indicate that its velocity is approximately zero and cross it out? I only say this because on a previous problem where I was asked to find the speed of a small object when it hits the Earth after being released from rest at a large height above the Earth's surface I used the same method and got the right answer without realizing that I didn't really set the problem up correctly. Any advice would be appreciated.

Thanks,
zachary570

You do realize that the velocities of the two masses are in inverse proportion to their respective masses?

Do you further see how this means that one can find a ratio between their kinetic energies?

Which gives you a way to answer the question that you asked. And gives you a way to solve the problem without asking that question in the first place.

jbriggs444 said:
You do realize that the velocities of the two masses are in inverse proportion to their respective masses?

Do you further see how this means that one can find a ratio between their kinetic energies?

Which gives you a way to answer the question that you asked. And gives you a way to solve the problem without asking that question in the first place.
Yes, I see what you mean and I see that ##\frac{K_{A}}{K_{B}} = \frac{m_{B}}{m_{A}}## and this allows me to solve for the velocity of A but this is just using the conservation of energy and doesn't seem any different from what I had done before. I guess what I don't understand is why when I find the velocity of a small mass when it hits the Earth starting from some height ##h##, the answer in the back of the book is $$v = \sqrt {2Gm_{E} \left( \frac{1}{R_{E}} - \frac{1}{R_{E} + h} \right)}$$ and not $$v = \sqrt \frac {{2Gm_{E} \left( \frac{1}{R_{E}} - \frac{1}{R_{E} + h} \right)}}{1 + \frac {m}{m_{E}}}$$ I get that the denominator is basically just 1 because of mass difference but I thought the answer came from ignoring the kinetic energy of the Earth and assuming it was 0.

zachary570 said:
I get that the denominator is basically just 1 because of mass difference but I thought the answer came from ignoring the kinetic energy of the Earth and assuming it was 0.
Aren’t they effectively the same approximation?
The small mass has KE
$$E_m = \frac {{Gmm_{E} ^2\left( \frac{1}{R_{E}} - \frac{1}{R_{E} + h} \right)}}{m + m_{E}}$$
Consider the Earth's KE:
$$E_E = \frac {{Gm ^2m_{E}\left( \frac{1}{R_{E}} - \frac{1}{R_{E} + h} \right)}}{m + m_{E}}$$
What's the ratio?

malawi_glenn and PeroK
haruspex said:
Aren’t they effectively the same approximation?
Ok, I see now. I didn't think to look at it that way. I see they boil down to the same ratio and understand why the ratio between the kinetic energies is the most important part. Thank you all!

@haruspex Yes. But do you know an intuitive explanation for why ##v_{1y}## is equal in modulus to ##u## for each angle? This seems surprising to me...

pepos04 said:
@haruspex Yes. But do you know an intuitive explanation for why ##v_{1y}## is equal in modulus to ##u## for each angle? This seems surprising to me...

## What is the principle of conservation of energy?

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In a closed system, the total energy remains constant over time.

## How is gravitational potential energy defined?

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is defined as the work done to move an object from a reference point to a specific position against the force of gravity, typically calculated as U = mgh, where U is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height above the reference point.

## How does gravitational potential energy change as an object moves?

As an object moves higher in a gravitational field, its gravitational potential energy increases. Conversely, as it descends, its gravitational potential energy decreases. This change in potential energy is often converted into kinetic energy or other forms of energy, maintaining the conservation of energy within the system.

## Can you give an example of conservation of energy involving gravitational potential energy?

Consider a pendulum. At its highest point, the pendulum has maximum gravitational potential energy and minimal kinetic energy. As it swings down, this potential energy is converted into kinetic energy. At the lowest point, the kinetic energy is at its maximum and the potential energy is at its minimum. As it swings back up, kinetic energy is converted back into potential energy. Throughout this motion, the total energy of the system remains constant, illustrating the conservation of energy.

## What happens to gravitational potential energy when an object falls freely under gravity?

When an object falls freely under gravity, its gravitational potential energy decreases as it loses height. This lost potential energy is converted into kinetic energy, increasing the object's speed as it falls. The total mechanical energy (sum of potential and kinetic energy) of the object remains constant throughout the fall, assuming no air resistance or other external forces act on the object.

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