The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius is 6.38 × 106 m, what is HST’s tangential speed?
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Answered by
0
Thats simple....
As per newton's law of gravitation
F = GMm/(d^2)
Where G = 6.67 * 10^(-11) Nm^2kg^(-2) , d is the central distance in metres and M and m are masses of two bodies
now, F = (6.67*10^(-11) * 11100 * 5.97*10^24)/(6.38 * 10^6 + 612000)^2
(Radius of earth + altitude of telescope = central distance)
=90410.79843 N
hence, the earth exerts a pull of 90410.79843 N on the telescope...
Hope I helped :)
Answered by
13
Answer:
7,570 m/s
Explanation:
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