The value of g a height of 100km from the surface of the Earth is nearly -Radius of the Earth - 6400km- -g on the surface of the Earth - 9-8m-s-2-9-5 ms-2-8-5 ms-2-10-5 ms-2-9-8 ms-2-

Question

Toppr Admin · Accepted Answer

Acceleration due to gravity changes with the height from Earth-apos-s surface as-g-x2032-g-1-x2212-2hR-x21D2-g-x2032-9-8-1-x2212-2-xD7-1006400-x21D2-g-x2032-9-8-1-x2212-132-9-49m-s2

The value of $g$ at a height of $100 k m$ from the surface of the Earth is nearly (Radius of the Earth $=$ 6400km) ( $g$ on the surface of the Earth $=$ $9.8 m / s^{2}$ )
$9.5 m s^{- 2}$
$8.5 m s^{- 2}$
$10.5 m s^{- 2}$
$9.8 m s^{- 2}$

Acceleration due to gravity changes with the height from Earth's surface as:
$g^{^{'}} = g (1 - \frac{2 h}{R})$
$\Rightarrow g^{^{'}} = 9.8 (1 - \frac{2 \times 100}{6400})$
$\Rightarrow g^{^{'}} = 9.8 (1 - \frac{1}{32}) = 9.49 m / s^{2}$

The value of g at a height of 100km from the surface of the Earth is nearly (Radius of the Earth = 6400km) (g on the surface of the Earth = 9.8m/s2)9.5ms−28.5ms−210.5ms−29.8ms−2

Acceleration due to gravity changes with the height from Earth's surface as:g′=g(1−2hR)⇒g′=9.8(1−2×1006400)⇒g′=9.8(1−132)=9.49m/s2

The value of $g$ at a height of $100 k m$ from the surface of the Earth is nearly (Radius of the Earth $=$ 6400km) ( $g$ on the surface of the Earth $=$ $9.8 m / s^{2}$ )
$9.5 m s^{- 2}$
$8.5 m s^{- 2}$
$10.5 m s^{- 2}$
$9.8 m s^{- 2}$

Acceleration due to gravity changes with the height from Earth's surface as:
$g^{^{'}} = g (1 - \frac{2 h}{R})$
$\Rightarrow g^{^{'}} = 9.8 (1 - \frac{2 \times 100}{6400})$
$\Rightarrow g^{^{'}} = 9.8 (1 - \frac{1}{32}) = 9.49 m / s^{2}$