Question

Planet X has 2 times the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitational field intensity at the surface of planet X? a) 9.8 m/s2 b) 4.9 m/s2 c) 19.6 m/s2 d) Sufficient data not given​

Answer 1

$\\\large{\underline{\underline{\sf{\pmb{\purple{Question:}}}}}}$

Planet X has 2 times the radius and 2 times the mass of the Earth. What is the best estimate for the magnitude of the gravitational field intensity at the surface of planet X?

• a) 9.8 m/s^2
• b) 4.9 m/s^2
• c) 19.6 m/s^2
• Sufficient data not given

$\\\large{\underline{\underline{\sf{\pmb{\purple{Required \: Answer:}}}}}}$

$\\ \underline{\underline{\sf{\tt{Given:}}}}$

• Radius of planet X = 2 times of the Earth
• Mass of the planet X = 2 times of the Earth.

$\\ \underline{\underline{\sf{\tt{To \: Find:}}}}$

• The gravitational field intensity at the surface of planet X

$\\ \underline{\underline{\sf{\tt{Solution:}}}}$

Let us assume that:-

• Mass of the Earth = m
• Mass of planet X = M
• Radius of the Earth = r
• Radius of planet X = R
• Gravitational field intensity of the Earth = g
• Gravitational field intensity of the Earth = g'

We know that:-

• $\underline{\boxed{\color{navy}{\rm{g = \dfrac{GM}{ {R}^{2}}}}}}$

Where,

• g = Gravitational field intensity of a planet
• G = Gravitational constant
• M = Mass of the planet
• R = Radius of the planet

According to the question:-

As for Earth:-

$\\ \sf{\blue{g = \dfrac{Gm}{ {r}^{2} } }} - - - - - - - - - - - (1)$

As for planet 'X'

$\\ \sf{\blue{g' = \dfrac{GM}{ {R}^{2} } }} - - - - - - - - - - - (2)$

We were given:-

• Radius of planet X = 2 times of the Earth

Therefore,

• M = 2m (M = Mass of planet X , m = Mass of the Earth)

Again,

• Mass of the planet X = 2 times of the Earth.

Therefore,

• R = 2r (R = Radius of planet X, r = Radius of the Earth)

Now, dividing equation (2) by equation (1) :-

$\\ \sf{\bold{ \dfrac{g'}{g} = \dfrac{GM}{ {R}^{2}} \div \frac{Gm}{ {r}^{2} } }}$

$\\ \sf{\implies{ \dfrac{g'}{g} = \dfrac{GM}{ {R}^{2}} \times \frac{{r}^{2}}{Gm } }}$

$\\ \sf{\implies{ \dfrac{g'}{g} = \dfrac{G \times 2 \times m}{ {(2 \times r)}^{2}} \times \dfrac{{r}^{2}}{Gm } }}$

$\\ \sf{\implies{ \dfrac{g'}{g} = \dfrac{\cancel{G} \times 2 \times \cancel{m}}{ {4\times \cancel{{r}^{2}} }} \times \dfrac{\cancel{{r}^{2}}}{\cancel{G}{\cancel{m}}}}}$

$\\ \sf{\implies{ \dfrac{g'}{g} = \frac{2}{4}}}$

$\\ \sf{\implies{ \dfrac{g'}{9.81m {s}^{ - 2} } = \frac{2}{4}}}$

$\\ \sf{\implies{ g' = \dfrac{2}{4} \times 9.81m {s}^{ - 2}}}$

$\\ \sf{\therefore{g' = \red{\frak{4.9m {s}^{ - 2}}}}}$

Hence, gravitational intensity of planet X is 4.9m/s^2

$\\\\\underline{\rm{Hence, option \: \bold{\green{b) 4.9 m/{s}^{2}} \: is \: the \: correct \: option.}}}$