Gravitational Potential Energy Project Thor

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  • #1
mailmas
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Homework Statement


Project Thor is a proposed (and terrifying) weapon system where a cylindrical tungsten rod (19600 kg m3 ) about the size of a telephone pole (6.10 m long and 0.300 m in diameter) is dropped from Earth orbit. Imagine you dropped one of these from an orbit 10,000 km above the surface of the earth, with what kinetic energy would it hit the ground? Assume no air resistance.

Homework Equations


m = p*Volume
KE - (mMG)/r+10000*1000 = 0
Radius Earth = 6.37 * 10^6
Mass Earth = 5.98 * 10^24

The Attempt at a Solution


m = 19600* *6.1 * (.15)^2 * 3.1415 = 8451
PE = (mMG)/r + 10000000
= ((8451)(5.98 * 10^24)(6.67 * 10^(-11)))/((6.37 * 10^6) +10000000)
= 205914267929J

 
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Answers and Replies

  • #2
Orodruin
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I am sorry, but it is utterly unclear what you are trying to do. For example, what does this
PE - (mMG)/r+10000*1000 = 0
mean? What is the 10000? What is the 1000? What makes you think this equation should hold? Also, please always write out your units. Newton is not a unit of energy.
 
  • #3
mailmas
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I am sorry, but it is utterly unclear what you are trying to do. For example, what does this

mean? What is the 10000? What is the 1000? What makes you think this equation should hold? Also, please always write out your units. Newton is not a unit of energy.
Got it.
PE = -(mMG)/r
PE_initial + KE_initial = PE_final + KE_final

PE_initial = KE_final
KE = -(mMG)/r

r = Radius Earth + 10,000km
= 6.37 * 10^6m + 10,000,000m

KE = -((8451kg * (5.98 * 10^24) kg * (6.67 * 10^(-11)) m^3/s^2*kg ) / ((6.37 * 10^6) + 10,000,000)m
 
  • #4
Orodruin
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PE_initial + KE_initial = PE_final + KE_final

PE_initial = KE_final
To get the second from the first you would need to assume that KE_initial = 0 and that PE_final = 0. This is not correct.
 
  • #5
mailmas
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To get the second from the first you would need to assume that KE_initial = 0 and that PE_final = 0. This is not correct.
Okay. So the KE_initial would look like:
KE = 1/2* m* v_tangental^2
v_tangental = sqrt(G*M/r)
KE = 1/2*m*G*M/r
 
  • #6
Orodruin
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Okay. So the KE_initial would look like:
No. If you do what the problem suggests and actually drop it from orbit, it will never fall. It will be in orbit! I believe the problem wants you to use an initial speed of zero. The problem is not in the kinetic energy. In indicating that you have zero potential energy at the end you are implicitly assuming that the Earth surface is the reference point for potential energy. Is this the case?
 
  • #7
mailmas
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No. If you do what the problem suggests and actually drop it from orbit, it will never fall. It will be in orbit! I believe the problem wants you to use an initial speed of zero. The problem is not in the kinetic energy. In indicating that you have zero potential energy at the end you are implicitly assuming that the Earth surface is the reference point for potential energy. Is this the case?
Yeah it would orbit if that is the case. And I used the center of mass as the reference point so my final potential energy would be mMG/Radius Earth?
 
  • #8
Orodruin
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And I used the center of mass as the reference point
No, you did not. Would the potential energy be zero at the center of mass if you used the expression ##V = -GMm/r##? (Note the minus sign, it is important!)
 
  • #9
Dr Dr news
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The potential energy is -GMm/r and the kinetic energy is GMm/2r so the total energy is -GMm/2r which is the energy required to leave an orbit and escape from the Earth's gravitational well to where the total energy is zero. To get an object to fall to the Earth, you would have to kill its velocity / kinetic energy in orbit. With no guidance your mass might wreck havoc but who knows where.
 

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