Question

If the radius of the earth shrinks by 1.5 % then the value of gravitational acceleration changes

Answer 1

Answer:

Given:

Radius of Earth shrinks by 1.5 %.

To find:

Change in the gravitational acceleration.

Concept:

Gravitational force

= (G × M × m)/r²

So we know acc = force/mass

=> Gravitational acceleration = force/mass

=> g = [ (G × M × m)/r² ] /m

=> g = GM/r² ,

where G is universal gravitational constant, M is mass of the Earth and r is the radius of the Earth.

Now , mass = volume × density.

So,

=> g = GM/r²

=> g = G (4/3 π r³ × ρ) /r²

=> g = 4/3 × π × G × ρ × r

So any small change in "r" will cause small change in "g" .

∆g/g = ∆r/r = 1.5%

So the change in gravity will be 1.5%

Answer 2

Question :-

Given above .

Answer :-

To find :-

→ Change of gravitational acceleration.

Explanation :-

We know what ,

gravitational force (f) is -

$\leadsto \sf \: (f) = \frac{G \: M \: m}{ {r}^{2} }$

• M is the mass of earth and r is the distance between earth and object .

• G is the universal gravitational constant .

We know that :

gravitational acceleration (g) is -

$\to \sf g \: = \frac{gravitational \: force}{mass} \\ \\ \to \sf \boxed{ \sf g = \frac{ G M\: }{ {r}^{2} } }$

• Earth is like a solid sphere hence its volume is -

$\to \: \sf \: v \: = \frac{4}{3} \pi {r}^{ 3}$

We know that :

→ density = mass/ volume

. • . mass (M)= density × volume

hence gravitational acceleration is -

$\to \sf \: g = \frac{G \{density( \rho) \times volume(v) \} }{ {r}^{2} } \\ \\ \to \sf g = \frac{ G \: \times \frac{4}{3} \pi \: {r}^{3} \times \rho}{ {r}^{2} } \\ \\ \to \sf \: \boxed{ \sf \: g \: = { \frac{4}{3} \times G \pi \: \times \: \rho \times \: r}}$

because r is change small hence small change in gravitational acceleration

$\to \sf \: \frac{ \Delta \: g\: }{g} = \frac{ \Delta \: r\: }{r} = 1.5 \:$

Hence 1.5 % radius shrinks of earth .