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

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of the Mars Climate Orbiter relative to Earth, as well as the potential energy of the Earth-spacecraft system. The subject area includes concepts from classical mechanics, specifically kinetic and potential energy calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to convert the spacecraft's speed from km/h to m/s and apply the kinetic energy formula. They express uncertainty about their calculations and question whether they are missing something. Other participants suggest rechecking arithmetic and clarify the need for a different approach to calculating gravitational potential energy.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on conversion methods and the appropriate equations for potential energy. There is a recognition of the need to reconsider assumptions regarding the formulas used, particularly for gravitational potential energy.

Contextual Notes

There is mention of potential confusion regarding the application of the gravitational potential energy formula, with participants noting that the standard mgh equation may not be applicable at the distances involved in this scenario.

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:
[tex]\frac{km}{hr} = \frac{1000m}{3600s}[/tex]
 
aww craters, the simplets thing you learn, you also forget.
 

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