Question

7. What will be the value of the acceleration due to gravity 'g' at a height 3200 km from the earth while its value at the earth is 9.8 m s ? Radius of the earth is 6400 km. Ans. 4.35 m s- -2​

Answer 1

Given :-

Radius of Earth = R = 6400 Km = 6.4 × 10⁶ m

Height = h = 3200 Km = 3.2 × 10⁶ m

Acceleration due to gravity on earth = 9.8 ms-²

Let, the acceleration due gravity on height = g'

As per relation we know that,

g' = g[R²/(R+h)²]

Putting the given values :-

g' = 9.8 [(6.4 × 10⁶)²/(6.4 × 10⁶ + 3.2 × 10⁶)²]

g' = 9.8[ 40.96 × 10¹²/92.16 × 10¹²]

g' = 9.8 × 0.44

g' = 4.34 ms-²

Answer 2

The general expression of Gravitational Acceleration at a height h from the Earth surface is given as :

$g_{2} = \dfrac{g}{ { \bigg( 1 + \dfrac{h}{R} \bigg)}^{2} }$

• 'h' is height , 'R' is radius of earth.

$\implies g_{2} = \dfrac{9.8}{ { \bigg( 1 + \dfrac{3200}{6400} \bigg)}^{2} }$

$\implies g_{2} = \dfrac{9.8}{ { \bigg( 1 + \dfrac{1}{2} \bigg)}^{2} }$

$\implies g_{2} = \dfrac{9.8}{ { \bigg( \dfrac{3}{2} \bigg)}^{2} }$

$\implies g_{2} = \dfrac{9.8 \times 4}{9 }$

$\implies g_{2} = 4.35 \: m {s}^{ - 2}$

So, acceleration at 3200 km height is 4.35 m/s².