Question

At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface ? ( Radius of Earth = 6400 km )

320 km

310 km

305 km

300 km ki l​

Answer 1

Question : At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface ? ( Radius of Earth = 6400 km )

a) 320 km

b) 310 km

c) 305 km

d) 300 km

Solution : Let, acceleration due to gravity above earth surface be ‘gh’ and radius of earth be ‘R’ and distance above the earth surface be ‘h’.

Now, according to question, here we know; distance above the surface of Earth the acceleration due to gravity decreases by 10 %.

Here, we know, acceleration due to gravity on the earths surface is denoted by g and g = 9.8 m/s²

Assume that x be the earths acceleration due to gravity decreases by 10 %.

Now,

⟹ {x/g} * 100 = 10 %

⟹ {x/9.8} * 100 = 10

⟹ x/9.8 = 10/100

⟹ x/9.8 = 0.1

⟹ x = 0.1 * 9.8

⟹ x = 0.98

Now,

⟹ gh = g - x

⟹ gh = 9.8 - 0.98

⟹ gh = 8.82 m/s²

Now, to find ‘ h ’ we use formula of acceleration due to gravity at height ‘ h ’.

⟹ gh = g * (1 - 2h/R)

⟹ 8.82 = 9.8 * (1 - 2h/6400)

⟹ 8.82/9.8 = 1 - 2h/6400

⟹ 0.9 = (6400 - 2h)/6400

⟹ 0.9 * 6400 = 6400 - 2h

⟹ 5760 = 6400 - 2h

⟹ - 2h = 5760 - 6400

⟹ - 2h = - 640

⟹ 2h = 640

⟹ h = 640/2

⟹ h = 320 km

Answer : Therefore, acceleration due to gravity is reduced by 10% at a distance of 320 km from the surface of the Earth. {( a ) 320 km}.