Question

If a body of mass m is taken out from a point below the surface of earth equal to half the radius of earth ,R, to a height R above the earth surface, then work done on it will be​

Answer 1

Answer:

The work done is W = (7/64) MgR

Explanation:

If a body of mass m is taken out from a point below the surface of earth equal to half the radius of earth, as being a huge amount of mass the mass of Earth also will be deducted ;

That's why the value of Gravitational acceleration (g) also will be changed.

Let Universal gravitational constant

(G) = 6.6743 × 10^-11 m^3 kg^-1 s^-2

Mass of Earth = M

Radius of Earth = R

Let, density of Earth = ρ

We know that :

$g = \frac{GM}{ {R}^{2} } \: \: ...equation(i)$

And we know that; Mass of Earth:

$M = \frac{4}{3} \pi \: {R}^{3} \rho$

So mass of the body having half the radius of earth = M` (Let):

$M \prime = \frac{4}{3} \pi \: {(\frac{R}{2}) }^{3} \\ = ( \frac{1}{8} ) \: \frac{4}{3} \pi \: {R}^{3} \\ = \frac{M}{8}$

So now the Mass of Earth is: M = M - M/8 = 7M/8

And now the decreased gravitational acceleration (g`) is :

$g \prime= \frac{G (\frac{7M}{8}) }{ {R}^{2} } \\ = ( \frac{7}{8}) \frac{GM}{{R}^{2}} \\ = \frac{7}{8}g$

So work done to take out this body to a height R above the earth surface (W) ;

(W) = F . R

$W = F . R \\ = M \prime \: g \prime \: . \: R \\ = ( \frac{M}{8}) \times ( \frac{7}{8} g) \times R \\ = \frac{7}{64} MgR$

Hence we get the work done is W = (7/64) MgR

Answer 2

Answer: If a body of mass m is taken out from a point below the surface of earth equal to half the radius of earth ,R, to a height R above the earth surface, then work done on it will be​ = $\frac{7}{64}Mgr$,

Acceleration due to gravity (g): The acceleration that is gained by an object as a result of the gravitational force acting on it is referred to as the acceleration due to gravity.

Explanation:

Step 1: The given data:-

The Universal gravitational constant

(G) = $6.6743 X 10^{-11} m^3 kg^{-1} s^{-2}$

Mass of Earth = M

Radius of Earth = R

Let, density of Earth = ρ

acceleration due to gravity when object is above the surface of Earth,

$g = \frac{GM}{R^2}$

Mass of the Earth if ρ is the density,

M = $\frac{4\pi R^3 \rho}{3}$

Step 2: The Decreased Mass of Earth:-

• If a body with a mass of m is removed from a position below the surface of the earth that is equivalent to half the radius of the earth, due to the fact that this is a significant amount of mass, the mass of the earth will also be subtracted

So mass of the body having half the radius of earth = M'

$M' = \frac{4 \pi \rho}{3}(\frac{R}{2})^3\\M'=\frac{M}{8}$

So now the Mass of Earth is: M = $M-\frac{M}{8}=\frac{7M}{8}$

Step 3: The decreased acceleration due to gravity (g'):-

$g'=\frac{GM'}{R^2}$

$g'=\frac{7GM}{8R^2}$

$g'=\frac{7}{8}g$

Step 4: Find the work done:-

To a height R above the earth surface, let W be the work done.

W = F.R

W = M'g'R

W = $\frac{M}{8}\frac{g}{8}R$

W = $\frac{7}{64}MgR$

Therefore the work done on it will be​   $\frac{7}{64}MgR$

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