Question

The value of acceleration due to gravity at Earth’s surface is 9.8 ms⁻². The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms⁻², is close to ______.

(Radius of earth = 6.4 × 10⁶ m)

Options

• 2.6 × 10⁶ m

• 6.4 × 10⁶ m

• 9.0 × 10⁶ m

• 1.6 × 10⁶ m

Answer 1

The value of acceleration due to gravity at Earth’s surface is 9.8 ms⁻². The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms⁻², is close to 2.6 × 10⁶ m.

Explanation:

Given: Acceleration due to gravity at Earth’s surface (g) = 9.8 ms⁻²

Acceleration due to gravity at altitude (g') = 4.9 ms⁻² = `1/2 g`

Radius of Earth (R) = 6.4 × 10⁶ m

The acceleration due to gravity (g') at an altitude (h) is given by:

g' = `g (R/(R + h))^2`

⇒ `1/2 g = g (R/(R + h))^2`

⇒ `1/2 = (R/(R + h))^2`    ...[Taking square root on both sides]

⇒ `1/sqrt 2 = R/(R + h)`

⇒ R + h = `sqrt 2 R`

⇒ h = `R (sqrt 2 - 1)`    ...[`sqrt 2` = 1.414]

= 6.4 × 10⁶ × (1.414 − 1)

= 6.4 × 10⁶ × 0.414

= 2.6496 × 10⁶

≈ 2.6 × 10⁶ m