A satellite in a circular orbit around earth with a radius 1.019 times the mean radius of the earth is hit by an incoming meteorite. A large fragment m=60 kg is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 373 m/s.

Question

A satellite in a circular orbit around earth with a radius 1.019 times the mean radius of the earth is hit by an incoming meteorite. A large fragment m=60 kg is ejected in the backwards direction so that it is stationary with respect to the earth and falls directly to the ground. Its speed just before it hits the ground is 373 m/s.

A. Find the total work done by gravity on the satellite fragment.

B. Find the amount of that work converted to heat

Thanks for the help on the water flow problem. I thought that we had to use the given velocity so no problems there. This problem is really giving me trouble I assume that I'm supposed to use PE=-G(Me)(m)/R to start but not sure

physics Bob Pursley plz help - bobpursley, Monday, November 30, 2009 at 5:05pm
Find the total PE available

PE=GMm/Re (1/2.019 -1/1)

Now, that has to be the work done
heat+finalKE=work done

ok so I posted this a couple days ago and I'm still having trouble. This is what I am doing

PE=(GMm/1.019Re)+(GMm/Re)

please help I'll be all done with physics for the semester if I can just finish this one problem!

Answer 1

changeofPE=(GMm/1.019Re)+(GMm/Re)

is correct. Now subtract the final KE from that, and then the remainder must be the heat given off.

Answer 2

To find the total work done by gravity on the satellite fragment, you need to calculate the gravitational potential energy (PE) it had before being hit by the meteorite.

The gravitational potential energy of an object is given by PE = -G(Me)(m)/R, where G is the gravitational constant, Me is the mass of the Earth, m is the mass of the object, and R is the radius from the center of the Earth to the object.

Since the satellite is in a circular orbit around the Earth, it can be assumed that its initial PE is equal to its initial kinetic energy (KE), as they are both constant throughout the orbit.

So, the total initial energy of the satellite is the sum of its initial potential energy and kinetic energy, given by PE + KE.

However, since the satellite is hit by the meteorite and a fragment is ejected, we need to find the change in its energy.

The change in energy is given by the work done on the satellite, which is equal to the final energy minus the initial energy.

Since the fragment falls directly to the ground and becomes stationary with respect to the Earth, its final kinetic energy is zero. Therefore, the final energy is just its final potential energy.

Now, let's solve the problem step by step:

A. Finding the total work done by gravity on the satellite fragment:

1. Find the initial potential energy (PE_initial) of the satellite:
PE_initial = -G(Me)(m)/(1.019Re), where Re is the mean radius of the Earth.

2. Find the final potential energy (PE_final) of the fragment:
PE_final = -G(Me)(m)/Re

3. The total work done by gravity is the change in potential energy:
Work = PE_final - PE_initial

B. Finding the amount of work converted to heat:

1. The work done by gravity is equal to the amount of energy converted to heat plus the final kinetic energy:
Work = Heat + Final KE

2. Since the fragment becomes stationary with respect to the Earth, its final kinetic energy is zero:
Final KE = 0

3. Therefore, the amount of work converted to heat is equal to the total work done by gravity:
Heat = Work

By following these steps, you can calculate the total work done by gravity on the satellite fragment and determine the amount of that work converted to heat.