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Q.

Consider a planet in the solar system which has a mass double the mass of the Earth and a density equal to the average density of the Earth. For an object weighing $W$ on the Earth, what will be the weight of the same object on the planet?

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a

2 W

b

3 W

c

W

d

2^{\frac{1}{3}} W

answer is D.

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Detailed Solution

A planet in the solar strength which has a mass doubled the mass of the Earth and density equal to the average density of the Earth. An object weighing

W

on the Earth and its weight on the planet will be

2^{\frac{1}{3}} W

.
The weight of an object is the product of its mass and acceleration due to gravity. For an object, the mass is constant, and the acceleration due to gravity varies.
The acceleration due to gravity varies as

g_{e} = \frac{G M_{e}}{R_{e}^{2}}

M_{e}

= mass of the earth

R_{e}

radius of earth

G

= universal gravitational constant
Given that mass of planet

M_{p} = 2 M_{e}

Mass of an object = Volume of the object \times Density of the object

Mass of an object = \frac{4}{3} π r^{3} \times ρ

Given that both planet and earth has same density

ρ

Mass of planet

M_{p}

is equal to double times of mass of earth

M_{e}

\frac{4}{3} π R_{p}^{3} \times ρ = 2 \times \frac{4}{3} π R_{e}^{3} \times ρ

Here

R_{p}

and

R_{e}

is the radius of planet and earth respectively.
Then,

r_{p} = 2^{\frac{1}{3}} R_{e}

Now the acceleration due to gravity of planet

g_{p} = \frac{G M_{p}}{R_{p}^{2}}

g_{p} = \frac{G \times 2 M_{e}}{{(2^{\frac{1}{3}} R_{e})}^{2}}

g_{p} = \frac{2}{2^{\frac{2}{3}}} \times g_{e}

g_{p} = 2^{\frac{1}{3}} \times g_{e}

The acceleration due to gravity of planet is

2^{\frac{1}{3}}

times of acceleration due to gravity

(g_{e})

of earth at the surface.
The weight of object on planet is

W_{P} = m g_{p}

W_{P} = m 2^{\frac{1}{3}} \times g_{e}

W_{p} = 2^{\frac{1}{3}} \times W_{e}

Hence, the weight of on object on planet is

2^{\frac{1}{3}}

times of weight on earth.

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