Physics, asked by jhamanita604, 21 days ago

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?​

Answers

Answered by lasthope71
0

Explanation:

Solution

q

=

R

2

GM

=

(4ve)

2

G×2me

=

8

1

ve

2

Gme

=

8

g

To find 4 times the acceleration due to gravity =4×

8

1

g=

2

g

=

2

10

=5m/s

2

Hence, the answer is 5m/s

2

.

Answered by XxDangerousQueenxX
3

 \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 }}}}}

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