Question

Value of g on the surface of earth is 9.8 m//s^(2). Find its value on the surface of a planet whose mass and radius both are two times that of earth.

Answer 1

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

As we know that :

$\large{\boxed{\boxed{\sf{g \: = \: G \dfrac{M}{R^2}}}}}$

Where,

• g is gravitational force
• G is gravitational constant
• M is mass of earth
• R is Radius of earth

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If a planet have Mass as well as Radius equal to double that of earth. Put R' = 2R

and M' = 2M

Substitute these values

$\implies {\sf{g' \: = \: G \dfrac{M'}{R'^2}}} \\ \\ \implies {\sf{g' \: = \: G \dfrac{2M}{(2R)^2}}} \\ \\ \implies {\sf{g' \: = \: G \dfrac{2M}{4R}}}$

Which is equal half of Gravitational force of earth. So,

$\implies {\sf{g' \: = \: \dfrac{g}{2}}} \\ \\ \implies {\sf{g' \: = \: g/2 \: = \: 4.6 \: N}}$

So, In this Planet value of g is also equal to the value of g on Earth which is 4.6 N

Answer 2

Answer:

Given:

Mass and radius of the planet is twice as compared to Earth.

To find:

Value of g on the surface of the planet.

Concept:

g or gravitational acceleration is actually a Gravitational Field Intensity. It's a vector Quantity denoting the strength of the field at a particular region wrt to the source.

Calculation:

Let Gravitational acceleration for earth be g .

$\huge{ \green{ \sf{g = G \dfrac{m}{ {r}^{2} }}}}$

For that planet :

mass = 2m

radius = 2r

$\huge{ \red{ \sf{g2 = G \dfrac{(2m)}{ {(2r)}^{2} }}}}$

$\huge{ \red{ \sf{ = > g2 = \dfrac{2}{4} \times \{ G \dfrac{m}{ {r}^{2} } \}}}}$

$\huge{ \red{ \sf{ = > g2 = \dfrac{1}{2} \times \{ G \dfrac{m}{ {r}^{2} } \}}}}$

$\huge{ \sf{ \red{ = > g2 = \dfrac{g}{2}}}}$

As per data , g = 9.8 m/s²

$\huge{ \sf{ \red{ = > g2 = \dfrac{9.8}{2}}}}$

$\huge{ \sf{ \red{ = > g2 = 4.9 \: m {s}^{ - 2} }}}$

So final answer :

$\boxed{ \huge{ \sf{ \blue{ g2 = 4.9 \: m {s}^{ - 2} }}}}$