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The mass of planet, and its diameter are three times those of earth. Then the acceleration due to gravity on the surface of the planet is:
1. 3.3 m/s2
2. 4.9 m/s2
3. 9.8 m/s2
4. 19.6 m/s2

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Correct Answer - Option 1 : 3.3 m/s2

The correct answer is option 1) i.e. 3.3 m/s2

CONCEPT:

  • Law of Universal Gravitation: It states that all objects attract each other with a force that is proportional to the masses of two objects and inversely proportional to the square of the distance that separates their centres.
    • It is given mathematically as follows

\(F = \frac{Gm_1m_2}{R^2}\)

Where m1 and m2 are the mass of two objects, G is the gravitational constant and R is the distance between their centres.

From the Law of Universal Gravitation, the gravitational acting on an object of mass m placed on the surface of Earth is

\(F = \frac{GMm}{R^2}\)

Where R is the radius of the earth. 

From Newton's second law, F = ma = mg

⇒ mg = \(\frac{GMm}{R^2}\)

Therefore, acceleration due to gravity, g = \(\frac{GM}{R^2}\)

CALCULATION:

Let acceleration due to gravity on Earth, g = \(\frac{GM}{R^2}\)

Let acceleration due to gravity on the planet be g'.

Given that:

Mass of the planet = 3 × Mass of earth ⇒ M' = 3 × M

Diameter of the planet = 3 × Diameter of earth ⇒ D' = 3 × D ⇒ Radius of planet, R' = 3 × Radius of earth = 3 × R

g'\(\frac{GM'}{R'^2} = \frac{G(3M)}{(3R)^2} \) 

\(\frac{1}{3}\frac{GM}{R^2}\) = \(\frac{1}{3}\) × 9.81 = 3.27 m/s2 ≈ 3.3 m/s2

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