A satellite used in a cellular telephone network has a mass of 2010kg and is in a circular orbit at a height of 770km above the surface of the earthTake the gravitational constant to be G = 6.67×10−11N⋅m2/kg2 , the mass of the earth to be me = 5.97×1024kg , and the radius of the Earth to be re = 6.38×106m .
What fraction is this of the satellite's weight at the surface of the earth?
Take the free-fall acceleration at the surface of the earth to be g = 9.80m/s2
2 answers
Tehe
(Gm1m2)/r^2
so
(6.67x10^-11)(2010)(5.97x10^24)/[(6.38x10^6)+770000]^2
note: 770km = 770000m
so
(6.67x10^-11)(2010)(5.97x10^24)/[(6.38x10^6)+770000]^2
note: 770km = 770000m
1 answer