Calculating Kinetic & Potential Energy of Rama's Impact

Click For Summary
SUMMARY

The discussion focuses on calculating the kinetic and potential energy of Rama, a fictional object from Arthur C. Clarke's "Rendezvous With Rama," with a mass of 907,184 kg and a velocity of 50,000 m/s just before impact with Earth. Participants confirm the use of the kinetic energy formula, Ek = 1/2mv^2, and the gravitational potential energy formula, E = Gm1m2/r, to determine the total energy at impact. The final equation for energy at impact is established as Efinal = Ekinetic + change of Epotential, emphasizing the need to account for both kinetic and potential energy changes during descent.

PREREQUISITES
  • Understanding of kinetic energy (Ek = 1/2mv^2)
  • Knowledge of gravitational potential energy (E = Gm1m2/r)
  • Familiarity with the concept of energy conservation
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research gravitational potential energy calculations in varying altitudes
  • Study the principles of energy conservation in physics
  • Explore advanced kinetic energy applications in astrophysics
  • Learn about the implications of mass and velocity on energy calculations
USEFUL FOR

Students studying physics, educators teaching energy concepts, and anyone interested in the application of energy equations in real-world scenarios.

Mecia22
Messages
9
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org


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 ?
 
Last edited:


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.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 3 ·
Replies
3
Views
994
Replies
55
Views
5K
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
6
Views
1K
Replies
15
Views
2K
Replies
6
Views
2K