A spherical planet has a radius of 7.2 \times 10^{6} \text{ m} and an acceleration due to … — JAMB UTME

A spherical planet has a radius of $7.2 \times 10^{6} \text{ m}$ and an acceleration due to gravity of $10 \text{ m/s}^2$ at its surface. What is the escape velocity for an object launched from the surface of this planet, in $\text{km/s}$ ?

Answer: 12 km/s

Answer

The escape velocity from the surface of the planet is 12 km/s.

The escape velocity

v_e

from the surface of a spherical planet of radius

R

with surface gravity

g

is given by

v_e = \sqrt{2gR}

. Substituting

g = 10 \text{ m/s}^2

and

R = 7.2 \times 10^6 \text{ m}

yields

v_e = \sqrt{2 \times 10 \times 7.2 \times 10^6} = \sqrt{1.44 \times 10^8} = 12000 \text{ m/s}

, which equals

12 \text{ km/s}

Step-by-Step Solution

Identify the relationship between surface gravity, planetary radius, and escape velocity

The escape velocity formula is

v_e = \sqrt{2gR}

Escape velocity is the minimum initial speed required for an object to overcome the gravitational pull of a celestial body.

Substitute the given numerical values into the formula

v_e = \sqrt{2 \times 10 \text{ m/s}^2 \times 7.2 \times 10^6 \text{ m}} = \sqrt{144 \times 10^6} \text{ m/s}

Plugging in the given values allows direct calculation of the velocity in standard SI units.

Simplify the square root and convert units to km/s

v_e = 12000 \text{ m/s} = 12 \text{ km/s}

Taking the square root of

144 \times 10^6

gives

12000 \text{ m/s}

, which corresponds to

12 \text{ km/s}

Key Concept

Escape Velocity from a Planet's Surface

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