How Do You Determine the Kinetic Energy of a Satellite in Orbit?

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To determine the kinetic energy of a satellite in orbit, the formula KE = 1/2 mv^2 is essential, where mass and speed are known. In the discussed example, a 100 kg satellite at 3500 km altitude has a kinetic energy of 30.3 GJ, calculated using its orbital speed of 22.9 x 10^3 km/h. The centripetal force equals gravitational force, allowing for the relationship v^2 = rg to be used in calculations. It is emphasized that once mass and speed are provided, the kinetic energy can be directly calculated without needing additional data. The discussion also highlights the importance of using consistent values for the radius of the Earth in calculations.
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1. A 100 kg satellite is placed in a circular orbit 3500 km above the surface of earth. At this elevation the acceleration of gravity is 4.09m/s_{2}. Determine the kinetic energy of the satellite , knowing that its orbitial speed is 22.9 x10_{3}km/h

The answer:30.3 GJ


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I have drawn the free- body diagram, but I still don't know how to approach this problem.
I have considerd that :
1) THere is no work done on satellite by the earth. since the satellite is in cicular motion
so, I cannot use W=Fs to consider ??
2) No PE change, No KE change
so, I cannot use TE=KE+PE to consider??
3) Finally, this is a gravitation problem??

I try my best to recall what I have learned in secondary school, however, I sill haven't any idea on this..

Please help
 
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If both mass and speed is given, then what's the problem in finding the KE? I think the speed should automatically come out from the other given data. So, I'll treat it as if you don't know thw speed.

The centripetal force should be equal to its weight.
So, mv^2/r = mg ==> v^2 = rg.
Now can you find mv^2/2?
 
Thanks for your reply, I can solve the problem by this method now
1) KE = 1/2 mv_{2}
where mass is known
2) F^{c}=F^{g}
\frac{mv_{2}}{r}= mg
3)the ration between the surface of Earth and at that specific height above the surface of the earth
g is inverse proportion to r_{2}

so I can get the answer correctly,
but I found another similar< actually the same> problem
and I use this method again, however i cannot get the right answer

Is above solution wrong?


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The similar problem is that:

1. A 400 kg satellite is placed in a circular orbit 6394km above the surface of earth. At this elevation the acceleration of gravity is 2.45m/s. Determine the kinetic energy of the satellite , knowing that its orbitial speed is 20000km/h

The answer:1.11MJ

so sorry that some mistakes are found on the first post


1. A 1500 kg satellite is placed in a circular orbit 3500 km above the surface of earth. At this elevation the acceleration of gravity is 4.09m/s^2. Determine the kinetic energy of the satellite , knowing that its orbitial speed is 22.9 x10^3km/h

The answer:30.3 GJ
 
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Please understand that KE is mv^2/2 and if the mass and speed both are given, you can find the KE simply by calculating mv^2/2. All the other data are irrelevant. So, the step 1 in your calc is enough. For your 1st prob, this gives a result of 32 KJ.

(For using Fc=mg, you have to assume some value for the radius of the earth. Taking 6370 km, I got KE = 2 GJ.)

Can you tell me in what context you obtained these problems?
 
THese problems are come from the reference book
"Vector Mechanics for Engineers - Dynamics" "Mc Graw - Hill"
 
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