If the acceleration due to gravity on the surface of the earth is $9.8 m / s^{2}$ , what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?

4.9 m / s^{2}

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9.8 m / s^{2}

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19.8 m / s^{2}

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None of the above

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Solution

The correct option is A $4.9 m / s^{2}$
$g \propto \frac{G M}{R^{2}}$
So, $g_{p} = (\frac{M_{p}}{M_{e}} \times (\frac{R_{e}}{R_{p}})^{2}) g$
$= 2 \times (\frac{1}{2})^{2} g$
$= \frac{g}{2} = 4.9 m / s^{2}$

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Solve the following

(1) If the weight of a body on the surface of the moon is 100N, what is its mass? (61.34 kg)

(2) If the acceleration due to gravity one the surface of the earth is 9.8 ms² what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth? (4.9 m/s²).

Q. Calculate the acceleration due to gravity at the surface of Mars if its diameter is 6760 km and mass one-tenth that of earth. The diameter of earth is 12742 km and acceleration due to gravity on earth is

9.8 m / s^{2}

Q. Value of

g

on the surface of earth is

9.8 m / s^{2}

. Find acceleration due to gravity at depth

h = \frac{R}{2}

from the surface (

R =

radius of earth).

Q. Imagine a planet

M

having mass same as the earth and having radius

3

times that of the earth.
a) Calculate the ratio of acceleration due to gravity on the surface of the planet

M

to the acceleration due to gravity on the surface of the earth.
b) If the average value of acceleration due to gravity on the surface of the earth is

9.8 {m s}^{- 2}

, then find the value of average acceleration due to gravity on the surface of planet

M

Q. Suppose a planet exist whose mass and radius both are half of those of earth, calculate acceleration due to gravity on the surface of this planet. g = 9.8 m/s^2

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If the acceleration due to gravity on the surface of the earth is 9.8m/s2, what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?

The correct option is A 4.9m/s2g∝GMR2So, gp=(MpMe×(ReRp)2)g=2×(12)2g=g2=4.9 m/s2

If the acceleration due to gravity on the surface of the earth is $9.8 m / s^{2}$ , what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?

The correct option is A $4.9 m / s^{2}$
$g \propto \frac{G M}{R^{2}}$
So, $g_{p} = (\frac{M_{p}}{M_{e}} \times (\frac{R_{e}}{R_{p}})^{2}) g$
$= 2 \times (\frac{1}{2})^{2} g$
$= \frac{g}{2} = 4.9 m / s^{2}$