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Energy conservation + planets

  • Thread starter NAkid
  • Start date
70
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1. Homework Statement
Two particles of masses M1 = 81000 kg and M2 = 9400 kg are initially at rest an infinite distance apart. As a result of the gravitational force the two masses move towards each other. Calculate the speed of mass M1 and mass M2 when their separation distance is 26.5 km.


2. Homework Equations



3. The Attempt at a Solution
I think you have to use energy-conservation equations here for each mass

1/2mv1^2 - (GM1M2)/R = 1/2mv2^2 - (GM1M2)/R

so, for mass M1, v1=0 and (GM1M2)/R on the left = 0 because R = infinite. then you get

1/2m1v2^2 = (GM1M2)/R --and solve for v2, but this doesn't seem to be correct..
 

Answers and Replies

1,166
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Lets say that the 2 masses were infinitely separated. The potential energy of the system would be 0J. However, an attractive force acts on them and they are brought closer together, in which the separation between them 26.5 km. Because it is an attractive force, the potential energy of the system is negative. What you do is find the change in potential energy. The initial value is 0J and the final value would be when they are the given distance apart. Using -dU = dK (d being delta, U being gravitational potential energy, and K being kinetic energy), you would solve for the kinetic energy of each mass. From that, you solve for velocity.
 
70
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isn't that basically what i did? calculated the changes to solve for velocity?
 
335
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Hint:

M1>>M2
 
454
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Hint:

M1>>M2
M1 / M2 <10 so I don't think you can say that.


-GM1M2/r is the potential energy of BOTH the masses.

At the start, potential and kinetic energy is 0. at a distance r the total energy is:

[tex](1/2)m_1v_1^2 + (1/2)m_2v_2^2 - Gm_1m_2/r[/tex]

This won't allow you to calculate [tex]v_1[/tex] and [tex]v_2[/tex] by itself. You also need conservation of momentum.
 
335
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M1 / M2 <10 so I don't think you can say that.


-GM1M2/r is the potential energy of BOTH the masses.

At the start, potential and kinetic energy is 0. at a distance r the total energy is:

[tex](1/2)m_1v_1^2 + (1/2)m_2v_2^2 - Gm_1m_2/r[/tex]

This won't allow you to calculate [tex]v_1[/tex] and [tex]v_2[/tex] by itself. You also need conservation of momentum.
I never told him to ignore the masses.

DID I?
 
454
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I never told him to ignore the masses.

DID I?
Well I didn't think you said that. I just don't think that M1>>M2 is either true or helpful.
 
335
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Well I didn't think you said that. I just don't think that M1>>M2 is either true or helpful.
It is helpful for sure, especially when energy consideration comes into play.
 
D H
Staff Emeritus
Science Advisor
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It is helpful for sure, especially when energy consideration comes into play.
It is not helpful for this problem. First, the approximation m1>>m2 is incorrect here, as m2 > m1/10. Second, the approximation implicitly makes the velocity of the larger mass zero, and the problem asks for both v1 and v2.

-GM1M2/r is the potential energy of BOTH the masses.

At the start, potential and kinetic energy is 0. at a distance r the total energy is:

[tex](1/2)m_1v_1^2 + (1/2)m_2v_2^2 - Gm_1m_2/r[/tex]

This won't allow you to calculate [tex]v_1[/tex] and [tex]v_2[/tex] by itself. You also need conservation of momentum.
This, on the other hand, is very good advice.
 

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