Question

A rocket is fired vertically with a speed of 5 km s^-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G = 6.67 × 10^–11 N m² kg^–2

Answer 1

Given:

Initial speed of rocket, v = 5 km s⁻¹ = 5 × 10³ m s⁻¹
Mass of Earth, M = 6.0 × 10²⁴ kg
Radius of Earth, R = 6.4 × 10⁶ m
Gravitational constant, G = 6.67 × 10⁻¹¹ N m² kg⁻²

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Step 1: Apply conservation of mechanical energy

The rocket is fired vertically upward and comes momentarily to rest at the maximum height.

Initial mechanical energy = Final mechanical energy

Initial energy at Earth’s surface:

E_i = (1/2)mv² − (GMm / R)

Final energy at maximum distance r from Earth’s center:

E_f = 0 − (GMm / r)

Equating E_i = E_f:

(1/2)mv² − (GMm / R) = − (GMm / r)

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Step 2: Simplify the equation

(1/2)v² = GM (1/R − 1/r)

Substitute values:

(1/2)(5 × 10³)² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴) (1/R − 1/r)

1.25 × 10⁷ = 4.002 × 10¹⁴ (1/R − 1/r)

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Step 3: Solve for r

1/R − 1/r = (1.25 × 10⁷) / (4.002 × 10¹⁴)

1/R − 1/r = 3.12 × 10⁻⁸

1/R = 1 / (6.4 × 10⁶) = 1.5625 × 10⁻⁷

1/r = 1.5625 × 10⁻⁷ − 3.12 × 10⁻⁸

1/r = 1.2505 × 10⁻⁷

r = 7.99 × 10⁶ m

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Step 4: Calculate maximum height above Earth’s surface

Maximum height, h = r − R

h = (7.99 − 6.4) × 10⁶

h = 1.6 × 10⁶ m

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Final Answer:

The rocket rises to a maximum height of
1.6 × 10⁶ m (≈ 1600 km) above the Earth’s surface before returning.