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Answer to Question #17498 in Mechanics | Relativity for Flora
Question #17498
If an object is near the surface of the earth, the variation of its weight with distance from the centre of the earth is negligible. The weight of an object at sea level is given as mg. Given that the acceleration due to gravity at sea level is 9.81 m/s2 and the radius of the earth is 6370km, at what height above the surface does the weight decrease to 0.99 mg? What will be the value of g at that height?
Expert's answer
F=G*m*M/r^2
G - gravitational constant
m - mass of body
M - mass of earth
F=G*m*M/r^2, where r=radius of earth=mg
So, G*m*M/r^2=mg
g=G*M/r^2
If g1=0.99g, then
0.99*G*M/r^2=G*M/r1^2
0.99/r^2=1/r1^2
r1=r/Sqrt(0.99)
So,
r1=6402.09
So, height is 32.09 m
g1=0.99g=9.7119 m/s^2
G - gravitational constant
m - mass of body
M - mass of earth
F=G*m*M/r^2, where r=radius of earth=mg
So, G*m*M/r^2=mg
g=G*M/r^2
If g1=0.99g, then
0.99*G*M/r^2=G*M/r1^2
0.99/r^2=1/r1^2
r1=r/Sqrt(0.99)
So,
r1=6402.09
So, height is 32.09 m
g1=0.99g=9.7119 m/s^2
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