The Gravitational Dance: Calculating Fred & Betty's KE

[rdn98]

Deep in space, two small asteroids wind up stationary, 1.7 km apart. Asteroid Fred has a mass of 14900 kg and Asteroid Betty has a mass of 16400 kg.
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a) Assuming the zero of gravitational potential energy to be when the asteroids are in contact, what is the potential energy of the two asteriod system Betty-Fred?
b) Betty and Fred start moving towards each other due to their mutual graviatation attraction. What is the total kinetic energy of the two asteriod system just before their collide?
c) What is the kinetic energy of asteroid Fred just before their collide?
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For a and b, its going to be the same answer. By using the U=Gm1m2/r equation, its giong to be the same energy. I mean, the potential energy gets converted entirely to kinetic energy. so the answer was 9.587E-6 J

now for part c, I am stuck. HOw do I go about breaking up that total Kinetic energy into two parts, which will ultimately help me find the answer for Fred?

 

[gnome]

Here's a clue:

Remember that since no external force is working on it, momentum for this system must be conserved. And since they start out stationary, the total momentum is always ... what?

 

[M98Ranger]

To find the kinetic energy of asteroid Fred just before they collide, we can use the conservation of energy principle. Since the total kinetic energy of the system is equal to the potential energy at the beginning, we can equate the two equations:

KE of system = PE of system

1/2mv^2 + 1/2mv^2 = Gm1m2/r

Simplifying, we get:

mv^2 = Gm1m2/r

Plugging in the values for mass and distance, we get:

14900v^2 = (6.67E-11)(14900)(16400)/(1.7E3)

Solving for v, we get:

v = 0.0078 m/s

Now, to find the kinetic energy of asteroid Fred, we can use the equation:

KE = 1/2mv^2

Plugging in the mass of Fred and the calculated velocity, we get:

KE = 1/2(14900)(0.0078)^2 = 0.457 J

Therefore, the kinetic energy of asteroid Fred just before they collide is 0.457 J.