find the height above the surface of earth where the value of acceleration due to gravity falls to 1% at the surface of the earth
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We know that gravity at any height is given as;
g’ = g (1 - 2h/R) —[1]
Decrease by 1% implies that g’ = 99/100g
Substitute In 1,we get
99/100g = g (1 - 2h/R)
99/100 = 1 - 2h/R
2h/R = 1 - 99/100
2h/R = 1/100
h = R/200
h = 6400/200 (R = 6400KM)
H = 32KM From surface of earth
g’ = g (1 - 2h/R) —[1]
Decrease by 1% implies that g’ = 99/100g
Substitute In 1,we get
99/100g = g (1 - 2h/R)
99/100 = 1 - 2h/R
2h/R = 1 - 99/100
2h/R = 1/100
h = R/200
h = 6400/200 (R = 6400KM)
H = 32KM From surface of earth
SisnuSivakumar:
it's wrong
acceleration due to gravity is just 1% on the surface
so what's the answer
57600 km
the radius of earth when g is equal to 1% is 64000km
the distance between present earth surface and new earth surface is equal to 64000km-6400km=57600km
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1
Answer:
Nice question . Answers will be 32km .
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