Solve Kepler's 3rd Law for Satellite Weight

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The discussion revolves around calculating the true weight of a satellite with a mass of 6000 kg when at rest on a planet's surface. Using Kepler's 3rd Law, the mass of the planet was determined to be 1.06 x 10^24 kg based on the satellite's orbit period and radius. The initial calculation for the planet's weight was incorrect due to using the satellite's height instead of the planet's radius. After correcting this, the weight of the satellite was recalculated to be approximately 25,235.69 N. The final confirmation of the calculation was deemed accurate by participants in the discussion.
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A satellite has a mass of 6000 kg and is in a circular orbit 4.40 10^5 m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.10 10^6 m. What is the true weight of the satellite when it is at rest on the planet's surface?

m=6000kg
t=7200 s

Mp = 4(3.14)^2 (4.40 * 10^5 + 4.10 * 10^ 6)^3/(6.67*10^-11)(7200)^2
Mp = 1.06 * 10^24
then using Mp for Wp (weight of planet)

Wp= (6.67*10^-11)(1.06 * 10^24)(6000) / (4.40 * 10^5)^2
* I used Keplers 3rd law to get the Mass of planet and then applied it to Weight of planet from which I got 2.191 x 10^6 and it still marks it wrong
 
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You put in the height of the satellite instead of the radius of the planet.
 
ohhhhhh
 
so its (6.67*10^-11)(1.06*10^24)(6000) / (4.10*10^6)^2
which is = 2.523569 * 10^4 which is same as 25235.69 ... can you please just confirm it for me? I have only 1 chance left, and I dnt want to lose points. Thanks
 
Seems good to me.
 
thanx
 

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