Question

9. At what height above the Earth's surface is the acceleration due to gravity 1% less than its value at the surface. Radius of Earth is 6400 km. (Take (1 + x) = 1 - 2r when x << 1)​

Answer 1

Answer:

Hint – In this question use the concept that value of acceleration due to gravity (g) at any height h above the earth’s surface is related to the acceleration due to gravity on the earth’s surface as g′=gR2(R+h)2 where R is the radius of the earth. This will help approach the solution of this problem.

Step by step answer:

Let at height h acceleration due to gravity becomes 64% of g

Given radius of the earth = 6400 Km.

Acceleration due to gravity is g m/s2.

Let acceleration due to gravity at height h = g’.

Therefore, g’ = 64% of g.

⇒g′=64100(g)=0.64g

Now the acceleration due to gravity at any height above the earth surface is given as,

⇒g′=gR2(R+h)2, where R = radius of the earth

Now substitute the values in the above equation we have,

⇒0.64g=gR2(R+h)2

Now as we see g is cancel out from both of the sides so we have,

⇒0.64=R2(R+h)2

Now take square root on both sides we have,

⇒0.64−−−−√=R2(R+h)2−−−−−−−−√

⇒0.8=RR+h

Now substitute the value of the radius of the earth in the above equation we have,

⇒0.8=64006400+h

Now simplify this equation we have,

⇒0.8(6400+h)=6400

Now divide by 0.8 throughout we have,

⇒(6400+h)=64000.8

Now simplify this we have,

⇒(6400+h)=8000

⇒h=8000−6400=1600 km.

So at the height of 1600 Km the value of the g becomes 64%of the g.

So this is the required answer.

Hence option (A) is the correct answer.

Note – A very interesting fact about acceleration due to gravity is that the force of gravity onto the equator is smaller as compared to the force an object experiences at the poles. The value of acceleration due to gravity is maximum at poles. This happens because g has inverse square dependence over the R that is the distance from the center of the earth as this distance changes and so does the value of g .