Energy of Rama just before hitting Earth

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Mecia22
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Homework Statement


-- This is out of Rendezvous With Rama by Arthur C. Clarke
You know it is 907184 kg and is going 50,000 m/s
So how much energy does it have right before it hits the earth?

Homework Equations



Do I use Fg= Gm1m2/r^2 ?
Ek=1/2mv^2?
I have ΔE=1/2mv^2 written down as an equation to use as well.

The Attempt at a Solution


I think I'm supposed to calculate the potential energy at 2 different spots and find the ΔE, but I'm not sure.

E=Gm1m2/r+1/2mv^2 would that work?
 
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At any particular point it has both KE and PE. As it descends, PE decreases and KE increases by the same amount. At the point you describe the KE is given by Ek=1/2mv^2, but to know what the energy will be when it strikes Earth you also need to know how much PE it will lose in the remaining descent. So you need the altitude at the point where the speed is known.
 


haruspex said:
At any particular point it has both KE and PE. As it descends, PE decreases and KE increases by the same amount. At the point you describe the KE is given by Ek=1/2mv^2, but to know what the energy will be when it strikes Earth you also need to know how much PE it will lose in the remaining descent. So you need the altitude at the point where the speed is known.


Okay, so then the equation would be E=EK+EP?
 


Depends what you mean by EP there. Write the equation in terms of four entities: Eki = KE at known initial point, Epi = PE at known initial point, Ekf = KE on collision, Epf = PE on collision.
 


Yikes... you're confusing me.
I guess what I'm asking is what steps would I do to get to my final equation?Would it be Efinal=Ekinetic+change of Epotential ?
 
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Mecia22 said:
Yikes... you're confusing me.
I guess what I'm asking is what steps would I do to get to my final equation?


Would it be Efinal=Ekinetic+change of Epotential ?

Yes, provided (i) Efinal means the KE on impact, and (ii) you get the sign right on the change in potential.