Kinetic energy of a spacecraft

  • Thread starter badman
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  • #1
badman
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i cant seem to figure this problem out.


Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87 \times 10^{6} \;{\rm km} from the earth and traveling at 1.20 \times 10^{4} \;{\rm km}/{\rm h} relative to the earth.


At this time, what was the spacecraft's kinetic energy relative to the earth?

i tried converting the traveling speed to meters per second, sqauring it, multiplying with the mass times 0.5 and i got it wrong. am i missing something here?


heres the second question What was the potential energy of the earth-spacecraft system?
is it asking me to find the gravitational potential energy or just the potential energy, using m*g*y=U?
 

Answers and Replies

  • #2
Doc Al
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badman said:
i tried converting the traveling speed to meters per second, sqauring it, multiplying with the mass times 0.5 and i got it wrong. am i missing something here?
That method is correct. Recheck your arithmetic.


heres the second question What was the potential energy of the earth-spacecraft system?
is it asking me to find the gravitational potential energy or just the potential energy, using m*g*y=U?
You need to find the gravitational potential energy, but you can't just use mgh. That equation is only good near the earth's surface. Find another equation for the potential energy between two masses. (Hint: The PE will depend on the distance between the spacecraft and the center of the Earth.)
 
  • #3
badman
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yeah i figured the equation for the gravitational energy.
 
  • #4
badman
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this is my calculation 1/2 times 629 kg times 1.20E4( in m/s its 2.0E5m/s)^2
 
  • #5
Doc Al
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Redo your conversion from km/hour to m/s:
[tex]\frac{km}{hr} = \frac{1000m}{3600s} [/tex]
 
  • #6
badman
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aww craters, the simplets thing you learn, you also forget.
 

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