Conservation of Energy with Gravitational Potential Energy

[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

 

[jbriggs444]

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.

 

[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.

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.

 

[haruspex]

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?

 

[zachary570]

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!

 

[pepos04]

@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...

 

[haruspex]

@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...

You've posted this in the wrong thread. Please move it.