Question

If the earth shrinks 50% of its real radius its mass remaining the same the weight of a body on the earth will
Explain this answer is decrease by 25 percentage is it true

Answer 1

Answer:

Given:

Radius of Earth decreases by 50%. But mass remains constant.

To find:

If there will be decrease in weight of a body by 25%. (True or False)

Concept :

Gravitational acceleration is actually a field intensity .

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \red{ \large{g = \dfrac{GM}{ {r}^{2} }}}}}$

Calculation:

Now, new mass is still M , but

new radius is r/2 (because radius decreases by 50%)

Let new gravity be g2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \red{ \large{g2 = \dfrac{GM}{ { (\frac{r}{2} )}^{2} }}}}}$

$\: \: \: \: \: \: \: \: { \sf{ \red{ \large{ \implies g2 = 4\dfrac{GM}{ { r}^{2} }}}}}$

$\: \: \: \: \: \: \: \: \: { \red{ \sf{ \large{ \implies g2 = 4g}}}}$

Since gravitational acceleration becomes 4 times , so the weight of any body on the surface of the Earth will also become 4 times.

So the given statement is false.

Answer 2

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Given :

• New Radius of earth is reduced 50%

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To Find :

• New Weight

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Solution :

We have formula for Acceleration due to y:

$\large{\boxed{\sf{g \: = \: G \dfrac{M}{R}}}}----(1)$

Where,

G is Gravitational Constant

g is gravitational force

M is mass of earth

R is radius of earth

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Now, If radius is reduced to 50% then. Put R' = R/2

$\large {\boxed{\sf{g' \: = \: G \dfrac{M}{R'}}}} \\ \\ \implies {\sf{g' \: = \: \dfrac{M}{\dfrac{R}{2}}}} \\ \\ \implies {\sf{g' \: = \: 4G \dfrac{M}{R}}} \\ \\ \implies {\sf{g' \: = \: 4g}}$