What Is the Kinetic Energy of the Mars Climate Orbiter Relative to Earth?

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SUMMARY

The kinetic energy of the Mars Climate Orbiter, with a mass of 629 kg and traveling at 12,000 km/h (3,333.33 m/s), can be calculated using the formula KE = 0.5 * m * v². The correct kinetic energy is approximately 10,500,000 J. For potential energy, the gravitational potential energy between two masses must be calculated using the formula PE = -G * (m1 * m2) / r, where r is the distance from the center of the Earth, not just mgh. The discussion emphasizes the importance of correct unit conversion and the appropriate equations for calculating energy in space contexts.

PREREQUISITES
  • Understanding of kinetic energy calculations using KE = 0.5 * m * v²
  • Familiarity with gravitational potential energy concepts and formulas
  • Knowledge of unit conversions, particularly from km/h to m/s
  • Basic grasp of Newton's law of universal gravitation
NEXT STEPS
  • Learn about gravitational potential energy calculations using PE = -G * (m1 * m2) / r
  • Study unit conversion techniques, especially for speed and distance in physics
  • Explore the implications of kinetic energy in orbital mechanics
  • Investigate the physics of spacecraft dynamics and energy calculations in space missions
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in understanding energy calculations related to spacecraft and orbital mechanics.

badman
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i can't 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?
 
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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.)
 
yeah i figured the equation for the gravitational energy.
 
this is my calculation 1/2 times 629 kg times 1.20E4( in m/s its 2.0E5m/s)^2
 
Redo your conversion from km/hour to m/s:
\frac{km}{hr} = \frac{1000m}{3600s}
 
aww craters, the simplets thing you learn, you also forget.
 

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