Question

A planet 'x' has mass 2 times and radius 3 times than that of Earth. What is the acceleration due to gravity on the planet,if acceleration due to gravity on Earth is 10m/s?

koi hai ?​

Answer 1

${q}^{i} = \frac{gm}{(4ve^{2}) } \\ \\ = \frac{1}{8} \frac{g \times 2me}{ve ^{2} } \\ \\ = \frac{g}{8} \\ \\ to \: find \: 4 \: times \: the \: \\ acceleration \: due \: to \: gravity \: \\ 4 \times \frac{1}{8}g = \frac{g}{2} = \frac{10}{2} \\ \\ = 5m |s ^{2} \\ \\ hence \: the \: answer \: s \: 5m |s ^{2}$

hope it will help u

Answer 2

$\large \dag$ Question :-

A planet 'x' has mass 2 times and radius 3 times than that of Earth. What is the acceleration due to gravity on the planet,if acceleration due to gravity on Earth is 10 m/s² ?

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Acceleration   \:  is \:  2.22\: m/s^2}} }$

$\large \dag$ Step by step Explanation :-

Let for Earth ;

Mass = M

Radius = R

Now let for planet 'x' ;

Mass = M'

Radius = R'

As per the question :-

✧ Mass of planet 'x' is two times the mass of earth ;

$\purple{ \large :\longmapsto  \underline { \pmb{\boxed{{\sf M'=2M} }}}}----(1)$

✧ Radius of planet 'x' is three times the radius of earth ;

$\purple{ \large :\longmapsto  \underline { \pmb{\boxed{{\sf R'=3R} }}}}----(2)$

❒ We know that acceleration due to gravity is :-

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{   \blue{ g= \small\frac{GM}{R {}^{2} } }}}}$

where,

G = Universal gravitational constant

M = Mass of planet

R = Radius of planet

Therefore for Earth ;

$:\longmapsto \rm g= \frac{GM}{R {}^{2} }$

and Given in question acceleration due to gravity on Earth is 10 m/s²,

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf  \small\frac{GM}{R {}^{2} } \large = 10 } }}}\small----(3)$

Now for planet 'x' ;

Acceleration due to gravity is :

$\: \: \: \: \rm g'= \frac{GM'}{{(R')}^{2} }$

⏩ Putting values of M' and R' from (1) and (2) :

$:\longmapsto \rm g'= \frac{G(2M)}{ {(3R)^{2}}}$

$:\longmapsto \rm g'= \frac{2GM}{9R {}^{2} }$

$:\longmapsto \rm g'= \frac{GM}{R {}^{2} } \times \frac{2}{9}$

⏩ Using (3) :

$:\longmapsto \rm g'=10 \times \frac{2}{9}$

$:\longmapsto \rm g'= \frac{20}{9}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf g' = 2.22 \: m/ {s}^{2} } }}}$

Therefore acceleration due to gravity on the planet 'x' is

$\large\underline{\pink{\underline{\frak{\pmb{\text Acceleration = 2.22 \: m/s^2 }}}}}$