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
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 calculate PE when R is (1/1.019)*3.38e6 and then calculate Pe when R is just 3.38e6? Or are you saying multiple (Pe)(1/1.019-1/1)?
To solve this problem, we can use the concept of work and energy.
A. To find the total work done by gravity on the satellite fragment, we can start by calculating the potential energy (PE) of the fragment at its initial position in the circular orbit. The gravitational potential energy can be calculated using the formula:
PE = -G * (Me) * (m) / R
Where G is the universal gravitational constant, Me is the mass of the Earth, m is the mass of the fragment, and R is the distance between the center of the Earth and the fragment.
But since the fragment is stationary with respect to the Earth and falls directly to the ground, its final potential energy is zero (as it is at the ground level). Therefore, the total work done by gravity on the fragment is equal to the change in potential energy:
Work done by gravity = PE_final - PE_initial
= 0 - PE_initial
= -PE_initial
So, the total work done by gravity on the satellite fragment is -PE_initial.
B. The amount of work converted to heat can be calculated using the work-energy principle. According to this principle, the work done on an object is equal to the change in its kinetic energy.
In this case, the fragment starts with zero velocity and ends with a speed of 373 m/s. Hence, its change in kinetic energy can be calculated as:
ΔKE = (1/2) * m * v^2
Where m is the mass of the fragment and v is its final velocity.
The amount of work converted to heat is equal to the total work done by gravity, which we already found to be -PE_initial. Therefore, we can equate the two:
-PE_initial = ΔKE
Now, we can substitute the formula for PE_initial (from part A) and solve for ΔKE:
-G * (Me) * (m) / R = (1/2) * m * v^2
By rearranging the equation, we get:
ΔKE = -2 * G * (Me) * (m) / R
So, the amount of work converted to heat is -2 * G * (Me) * (m) / R.
Remember to substitute the appropriate values for G, Me, m, and R to find the final numerical result.