Question

The radius of planet A is double the radius of
planet B. If the mass of A is MA, what must be
the mass of B so that the value of g on B is
double that of its value on A?
​

Answer 1

Given :

• The radius of planet A is double the radius of planet B
• Mass of A is $\sf{M_A}$

To Find :

• The mass of B so that the value of g on B is double that of it's value on A

Solution :

Acceleration due to gravity is given by ,

$\\ \star \: {\boxed{\sf{\purple{g = \dfrac{Gm}{ {R}^{2} } }}}}$

Where ,

• g is acceleration due to gravity
• m is mass of planet
• G is Gravitational constant
• r is radius of planet

Let ,

• The radius of B be R

We have ,

• Radius of A = 2R [given condition]
• Mass of A = $\sf{M_A}$
• Radius of B = R
• Mass of B = (let $\sf{M_B}$)

Calculating the value of g on planet A ,

$\\ : \implies \sf \: g_A = \dfrac{Gm_A}{ {R_A}^{2} }$

$\\ : \implies \sf \: g_A = \dfrac{GM_A}{ {(2R)}^{2} }$

$\\ : \implies \sf \: g_A = \dfrac{Gm_A}{4 {R}^{2} } \: ............(1)$

Now calculating the value of g on planet B ,

$\\ : \implies \sf \: g_B = \dfrac{Gm_B}{ {R_B}^{2} }$

$\\ : \implies \sf \: g_B = \dfrac{GM_B}{ {R}^{2} } \: ...........(2)$

According to the given question ,

$\longrightarrow\sf{2g_A=g_B}$

$\longrightarrow\sf{\dfrac{g_B}{g_A}=2}$

Now ,

$\\ : \implies \sf \: \dfrac{\frac{GM_B}{ {r}^{2} } }{ \frac{GM_A}{4 {r}^{2} } } = 2 = \frac{g_B}{g_A}$

$\\ : \implies \sf \: \frac{M_B}{ \frac{M_A}{4} } = 2$

$\\ : \implies \sf \: \frac{4M_B}{M_A} = 2$

$\\ : \implies \sf \: 4M_B = 2M_A$

$\\ : \implies \sf 2M_B = \:M_A$

$\\ : \implies{\underline{\boxed{\pink {\sf{ \:M_B = \frac{M_A}{2} }}}}} \: \bigstar$

Hence ,

• The Mass of the planet B should be Half the mass of planet A so that the value of g on B is double that of it's value on A

Answer 2

$\large{\boxed{\underline{\sf{Question}}}}$

The radius of planet A is double the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is double that of its value on A ?

$\large{\boxed{\underline{\sf{Given \: that}}}}$

✰ The radius of planet A is double the radius of planet B.

✰ The mass of planet A is $M_{A}$

$\large{\boxed{\underline{\sf{To \: find}}}}$

✰ The mass of planet B so that the value of g on B is double that of its value on A.

$\large{\boxed{\underline{\sf{Solution}}}}$

✰ The mass of planet B so that the value of g on B is double that of its value on A =

$\large{\boxed{\underbrace{\sf{Understanding \: the \: concept}}}}$

✠ This question says that there are 2 planet's namely A and B. The radius of planet A is double the radius of planet B. The mass of planet A is $M_{A}$ Afterthat this question ask us to find the mass of planet B so that the value of g on B is double that of its value on A ?

$\large{\boxed{\underbrace{\sf{Using \: concepts}}}}$

✰ Acceleration due to the gravity.

$\large{\boxed{\underbrace{\sf{Using \: formulas}}}}$

✰ Acceleration due to the gravity = ${\boxed{\sf{\green{g = \dfrac{Gm}{ {R}^{2} } }}}}$

$\large{\boxed{\underbrace{\sf{We \: also \: write \: these \: as}}}}$

✠ Mass of the body as m.

✠ Acceleration due to gravity as g.

✠ Gravitational constant as G

✠ Radius of planet as r.

$\large{\boxed{\underbrace{\sf{Assumption}}}}$

✠ Let radius of planet B be $R$

✠ Let mass of planet B be $M_{B}$

$\large{\boxed{\underbrace{\sf{Full \: solution}}}}$

✰ Let's see (finally) what is given again to understand the concept easily.

☀ Mass of planet A = $M_{A}$

☀ Radius of A planet = $2R$

☀ Mass of planet B = $M_{A}$

☀ Radius of planet B = $R$

_________________________________

Now, calculating the g on the planet A.

$\large{\bf{g_{a} = \frac{Gm_{A}}{R_{A}^{2}}}}$

$\large{\bf{g_{a} = \frac{Gm_{A}}{(2R)^{2}}}}$

$\large{\bf{g_{a} = \frac{Gm_{A}}{4R^{2}}}}$ ᗴquation ①

_________________________________

Now, calculating the g on the planet B.

$\large{\bf{g_{b} = \frac{Gm_{B}}{R_{B}^{2}}}}$

$\large{\bf{g_{b} = \frac{Gm_{B}}{(R)^{2}}}}$ ᗴquation ②

_________________________________

Now, according to the question.

$\large{\bf{2g_{A}}}$ = $\large{\bf{g_{B}}}$

$\large{\bf{\frac{g _{B}}{g_{A} = 2}}}$

$\large{\bf{\frac{ \frac{ \frac{Gm_{B}}{r^{2}}}{Gm_{A}}}{4r^{2}}}}$ = $\large{\bf{2 = \frac{G_{B}}{G_{A}}}}$

$\large{\bf{\frac{ \frac{M_{B}}{M_{A}}}{4 = 2}}}$

$\large{\bf{4M_{B}}}$ = $\large{\bf{M_{A}=2}}$

$\large{\bf{4M_{B}}}$ = $\large{\bf{2M_{A}}}$

$\large{\bf{2M_{B}}}$ = $\large{\bf{M_{A}}}$

$\large{\boxed{\rm{M_{B} = \frac{M_{A}}{2}}}}$