calculate acceleration due to gravity at a height of 3200 km from the Earth surface and (Me is equal to 6.0 into 10 ki power 24 kg and Re is equal to 6400 km)
Answers
Answer:Assume-4.355m/s^2 ~ 4.4m/s^2
h= 3200km = 3200000m = 3.2 x10^6
R= 6400km = 6400000m = 6.4 x 10^6
g= 9.8m/s^2
We know that-
g'= g[R^2/(R+h)^2]
= 9.8(40.96 x 10^12/92.16 x 10^12)
= 9.8 x 4/9
= 4.355m/s^2 ~ 4.4m/s^2
Thus, value of g at a place 3200km above surface of the earth is 4.355m/s^2 ~ 4.4m/s^2
Explanation::Assumeh= 3200km = 3200000m = 3.2 x10^6
R= 6400km = 6400000m = 6.4 x 10^6
g= 9.8m/s^2
We know that-
g'= g[R^2/(R+h)^2]
= 9.8(40.96 x 10^12/92.16 x 10^12)
= 9.8 x 4/9
= 4.355m/s^2 ~ 4.4m/s^2
Thus, value of g at a place 3200km above surface of the earth is 4.355m/s^2 ~ 4.4m/s^2
Explanation:
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