Gravitational Potential Energy question

In summary, in this conversation, the problem at hand involves a binary star system with two stars, each with the same mass as our sun, separated by 1.00×10^12 m. A comet, located very far away and essentially at rest, is being pulled towards the stars by gravity along a straight line that passes through the midpoint between the two stars. The question at hand is to determine the comet's speed at the midpoint. To solve this problem, the formula K_f + U_f = K_i + U_i is used, as energy is conserved. The final potential energy (U_f) is found to be 2 times the potential energy of one star, and the initial potential energy (U_i) is
  • #1
neo982
7
0
Homework Statement
A binary star system has two stars, each with the same mass as our sun, separated by 1.00×10^12 m. A comet is very far away and essentially at rest. Slowly but surely, gravity pulls the comet toward the stars. Suppose the comet travels along a straight line that passes through the midpoint between the two stars.

What is the comet's speed at the midpoint?

The attempt at a solution

I know that this is supposed to use the K_f +U_f = K_i + U_i formula because energy is conserved. But the hard part to this problem is the reasoning behind it. I thing that the final potential energy (U_F) should be zero there fore giving me the eq. K_f = K_i + U_i...but then I start to wonder if the K_i should be zero also because it says the velocity is essentially at zero. When trying some of these ideas I end up with a radius (for the potential energy formula) -Gmm/r also I am not sure how to factor in the mass of both the planets and the radius between them, I initially thought that I should treat them as one big planet with total mass. I am close, but just need help with the conceptual part of this problem.
 
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  • #2
neo982 said:
I think that the final potential energy (U_F) should be zero there fore giving me the eq.
Nopes, I don't think so
@ the midpoint it is r distance away from both stars

Therefore U_f = 2*(U with one star)

And using the formula GMm/r, you can find U_i (r is inf)

and K_f is just change in U


Edit: I think you are right for most of the part.. and in the end that should leave you with this equation:
K_f+U_f = 0
 
Last edited:
  • #3
Thanks I got it.. K_f + U_f = 0 ..therefore v_f = sqrt([2*G(2*mass of planet)/r]) where radius is 1/2 of the value the question gives..which is the diameter...which is 32,586 m/s

Thanks.
 

Related to Gravitational Potential Energy question

1. What is gravitational potential energy?

Gravitational potential energy is the energy stored within an object due to its position in a gravitational field. It is the potential for the object to do work as a result of its position.

2. How is gravitational potential energy calculated?

Gravitational potential energy is calculated by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s^2) and the height of the object above the ground.

3. What is the unit of measurement for gravitational potential energy?

The unit of measurement for gravitational potential energy is joules (J). This is the same unit used to measure other forms of energy.

4. How does gravitational potential energy affect an object's motion?

Gravitational potential energy does not directly affect an object's motion. However, as an object moves in a gravitational field, potential energy is converted into other forms of energy, such as kinetic energy, which can affect an object's motion.

5. Can gravitational potential energy be negative?

Yes, gravitational potential energy can be negative. This usually occurs when an object is below the reference level, such as when it is in a hole or underground. In these cases, the potential energy is measured relative to the reference level, and a lower position results in a negative value.

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