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Considering a planet whose mass and radius both are half that of the earth. What will be the acceleration due to gravity on the surface of this planet?

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Half

Double

Four times

Triple

answer is B.

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Detailed Solution

Considering a planet whose mass and radius both are half that of the earth. The acceleration due to gravity on the surface of this planet will be double.
Acceleration due to gravity of earth

g = G \times \frac{M_{e}}{R^{2}} = 9.8 m / s^{2}

Where

M_{e}

= mass of the earth

R

= Radius of the earth
If the mass of the planet,

m = M_{e} / 2

,
And radius of the planet

r = R / 2

Gravitational acceleration on the surface of that planet will be:

g = G \times \frac{m}{r^{2}}

……..(i)
⇒

m = \frac{M_{e}}{2}

………(ii)
⇒

r = \frac{R}{2} \dots \dots . . (

iii)
Put (ii) and (iii) in (i) eq. we get
⇒

g

G \times \frac{\frac{M}{2}}{{(\frac{R}{2})}^{2}} = \frac{4}{2} \times (G \times \frac{M}{R^{2}})

As we know

G \times \frac{M}{R^{2}}

= 9.8 m / s^{2}

⇒

g = 2 \times 9.8 m / s^{2}

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