If radius of earth is 6400 km,What will be the weight of 1 quintal body if taken to the height of 1600 km above the sea level?
Answers
Answered by
56
We know that,
Change in acceleration due to gravity by taking to a height h above the surface g'= g/( 1 +h/R)²
Now, Acceleration at height " 1600 km " = g/ ( 1 + 1600/6400 )² = g /( 1 +1/4)²= 16g/25 = 6.272 m/s²
We know that, The weight of the body is subjected as force on the earth.
So,
W = mg
Given, m = 100 kg.
Now, Weight " W " = mg = 100(6.272) = 627.2 N.
Hope helped!
Change in acceleration due to gravity by taking to a height h above the surface g'= g/( 1 +h/R)²
Now, Acceleration at height " 1600 km " = g/ ( 1 + 1600/6400 )² = g /( 1 +1/4)²= 16g/25 = 6.272 m/s²
We know that, The weight of the body is subjected as force on the earth.
So,
W = mg
Given, m = 100 kg.
Now, Weight " W " = mg = 100(6.272) = 627.2 N.
Hope helped!
Answered by
23
formula of acceleration due to gravity at h height from the surface of earth/sea level when radius of earth is R is given by
Here g₀ is acceleration due to gravity on sea level/earth surface
one thing we also know,
e.g., acceleration due to gravity on sea level/earth surface , g₀ = 9.8m/s²
So, g' = 9.8/(1 + 1600/6400)² [ ∵ h = 1600 km , R = 6400 km]
g' = 9.8/(1 + 1/4)² = 16 × 9.8/25 m/s²
Now, weight of body = mass of body × g'
= 100 × 16 × 9.8/25
= 64 × 9.8 = 627.2N
Here g₀ is acceleration due to gravity on sea level/earth surface
one thing we also know,
e.g., acceleration due to gravity on sea level/earth surface , g₀ = 9.8m/s²
So, g' = 9.8/(1 + 1600/6400)² [ ∵ h = 1600 km , R = 6400 km]
g' = 9.8/(1 + 1/4)² = 16 × 9.8/25 m/s²
Now, weight of body = mass of body × g'
= 100 × 16 × 9.8/25
= 64 × 9.8 = 627.2N
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