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JAMB Physics 2014 Paper

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Question : 6

Total: 41

Calculate the escape velocity of a satellite launched from the earth's surface if the radius of the earth is 6.4x10⁶m

25.3kms^-1
4.2kms^-1
4.0kms^-1
11.3kms^-1

Solution:

Concept:

Escape velocity is the minimum speed a body needs to leave Earth's gravitational field permanently.

Explanation:

Use the formula

v_{\text{esc}} = \sqrt{2gR}

, where

R

is Earth's radius and

g

is acceleration due to gravity.

Take

g \approx 10 \, \mathrm{m/s^2}

and

R = 6.4 \times 10^6 \, \mathrm{m}

.

Substitute the values:

v_{\text{esc}} = \sqrt{2 \times 10 \times 6.4 \times 10^6}

.

This simplifies to

v_{\text{esc}} = \sqrt{1.28 \times 10^8}

.

\sqrt{1.28 \times 10^8} = 11313.7 \, \mathrm{m/s}

.

Convert to

\mathrm{km/s}

:

11313.7 \div 1000 = 11.3 \, \mathrm{km/s}

.

Answer:

D.

11.3 \, \mathrm{km\,s^{-1}}

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