Question

Find out please????????????????????







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Answer 1

Answer:

Given:

Mass of planet is 20% more than that of Earth. The radius of that planet is 20% less than that of Earth.

To find:

Acceleration due to gravity on the planet

Calculation:

Gravitational acceleration is a Field Intensity vector directed towards the centre of the source.

Let gravitational force be F

$\boxed{ \sf{ \orange{F = \dfrac{GMm}{ {r}^{2} }}}}$

Now gravitational acceleration os force divided by mass of object (m)

$\boxed{ \sf{ \orange{g = \dfrac{GM}{ {r}^{2} }}}}$

Calculation:

Let mass of Earth be M and it's radius be r .

So, mass of planet is

$\sf{M2 = M \times \dfrac{120}{100}} = \dfrac{6M}{5}$

Radius of planet is

$\sf{r2 = r \times \dfrac{80}{100} = \dfrac{4r}{5} }$

Now gravitational acceleration in that planet :

$\sf{ \red{g2 = \dfrac{G( \dfrac{6M}{5} )}{ {( \dfrac{4r}{5} )}^{2} }}}$

$\sf{ \red{\implies g2 = \dfrac{GM}{ {r}^{2} } \times \dfrac{15}{8} }}$

$\sf{ \red{\implies g2 = 9.8 \times \dfrac{15}{8}}}$

$\sf{ \red{\implies g2 = 18.375 \: m {s}^{ - 2}}}$

So correct option is :

$\boxed{ \boxed{ \sf{ \large{ \bold{ \green{option \: b)}}}}}}$

Answer 2

$\underline{\boxed{\mathfrak{ \huge{ \orange{AnSwEr}}}}} \\ \\ \dagger \: \: \boxed {\blue{ \rm{Given}}}$

Mass of planet is 20% more than that of earth.

The radius os that planet is 20% less than that of earth.

$\dagger \: \: \boxed{ \rm{ \blue{To \: Find}}}$

Acceleration due to gravity of the planet

$\dagger \: \: \boxed{ \rm{ \blue{Formula}}}$

Gravitational acceleration of any planet is given as

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \rm{ \pink{g = \frac{GM}{ {R}^{2} }}}}}$

where, M = mass of planet and R = radius of planet

let

M1 = mass of earth

M2 = mass of planet

R1 = radius of earth

R2 = radius of planet

g1 = gravitational acc. of earth

g2 = gravitational acc. of planet

As per question....

$\rm \: M{ \tiny{2}} = M{ \tiny{1}} + \frac{20M{ \tiny{1}}}{100} = \frac{120M{ \tiny{1}}}{100} = \frac{6M{ \tiny{1}}}{5} \\ \\ \rm \: R{ \tiny{2}} = R{ \tiny{1}} - \frac{20R{ \tiny{1}}}{100} = \frac{80R{ \tiny{1}}}{100} = \frac{4R{ \tiny{1}}}{5} \\ \\ \dagger \: \: \boxed{ \rm{ \blue{Calculation}}}$

gravitational acc. of planet is given as

$\leadsto \rm \: g{ \tiny{2}} = \frac{GM{ \tiny{2}}}{ {R{ \tiny{2}}}^{2} } = \frac{g( \frac{6M{ \tiny{1}}}{5} )}{ ({ \frac{4R{ \tiny{1}}}{5} })^{2} } = \frac{15}{8} \frac{gM{ \tiny{1}}}{ {R{ \tiny{1}}}^{2} } \\ \\ \leadsto \rm \: g{ \tiny{2}} = \frac{15}{8} \times 9.8 = 18.375 \: \frac{m}{ {s}^{2} } \\ \\ \underline{ \boxed{ \rm{ \huge{ \orange{g{ \tiny{2}} = 18.375 \: \frac{m}{ {s}^{2} } }}}}} \: \: \red{ \huge{ \star}}$