A planet has a mass equal to 8 times the mass of the Earth and a radius equal to 2 times t… — JAMB UTME

A planet has a mass equal to $8$ times the mass of the Earth and a radius equal to $2$ times the radius of the Earth. If the acceleration due to gravity on the surface of the Earth is $10\text{ m/s}^2$ , what is the acceleration due to gravity on the surface of the planet in $\text{m/s}^2$ ?

Answer: 20 m/s^2

Answer

The acceleration due to gravity on the surface of the planet is

20\text{ m/s}^2

The acceleration due to gravity at the surface of a spherical body is given by

g = \frac{GM}{R^2}

. When the mass is multiplied by

8

and the radius is multiplied by

2

, the new acceleration becomes

g_p = \frac{G(8M)}{(2R)^2} = \frac{8}{4}\frac{GM}{R^2} = 2g

. Substituting

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

gives

20\text{ m/s}^2

Step-by-Step Solution

Write the expression for acceleration due to gravity at the surface of Earth.

g_e = \frac{GM_e}{R_e^2} = 10\text{ m/s}^2

Gravitational field strength at the surface depends directly on body mass and inversely on the square of radius.

Set up the ratio equation for the planet's gravitational acceleration using relative mass and radius values.

g_p = \frac{G(8M_e)}{(2R_e)^2} = \frac{8}{4} \left(\frac{GM_e}{R_e^2}\right) = 2g_e

Squaring the radius multiplier of

2

yields

4

in the denominator, while the numerator increases by a factor of

8

Calculate the final numeric value.

g_p = 2 \times 10\text{ m/s}^2 = 20\text{ m/s}^2

Multiplying Earth's value by the combined ratio of

2

gives the answer.

Key Concept

Gravitational field strength on planetary surfaces (

g = \frac{GM}{R^2}

)

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Key Concept

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