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Keplers 3rd Law

  • Thread starter ny_aish
  • Start date
  • #1
21
0
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
 

Answers and Replies

  • #2
430
2
You put in the height of the satellite instead of the radius of the planet.
 
  • #3
21
0
ohhhhhh
 
  • #4
21
0
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
 
  • #5
430
2
Seems good to me.
 
  • #6
21
0
thanx
 

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