Question

Consider an object of mass $\mathit{m}$, not necessarily small compared with the mass of the Earth, released at a distance of $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathit{m}$from the center of the Earth. Assume the Earth and the object behave as a pair of particles, isolated from the rest of the Universe. (a) Find the magnitude of the acceleration ${\mathit{a}}_{\mathbf{r}\mathbf{e}\mathbf{l}}$with which each starts to move relative to the other as a function of $\mathit{m}$. Evaluate the acceleration (b) for $\mathit{m}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathit{k}\mathit{g}$(c) for $\mathit{m}\mathbf{=}\mathbf{200}\mathit{k}\mathit{g}$, and (d) for $\mathit{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{24}}\mathbf{}\mathit{k}\mathit{g}$. (e) Describe the pattern of variation of ${\mathit{a}}_{\mathbf{r}\mathbf{e}\mathbf{l}}$with$\mathit{m}$ .

Short answer

The relation of the relative acceleration is found to be $a_{rel}=\frac{G{M}_E}{r^2}\left(1+\frac{m}{M_E}\right)$

Step-by-step solution

Given Data.

The distance of the object from the earth is$r=1.20\times {10}^{7}m$ .

The mass of the earth is$M_E=5.97\times {10}^{24}kg$ .

The mass of the object for subpart (b) is$m=5.00kg$ .

The mass of the object for subpart (c) is $m=2000kg$.

The mass of the object for subpart (d) is$m=2.00\times {10}^{24}kg$ .

Gravitational field

The force on an object by Earth divided by the mass of the object gives us the gravitational field on the object.

$\begin{array}{rcl}g& =& \frac{F_g}{m}\\ & =& \frac{\frac{G{M}_{E}m}{r^2}}{m}\\ g& =& \frac{G{M}_E}{r^2}\\ & & \end{array}$

Where,$r$ is the distance of the object from the centre of the earth. This is also called acceleration due to gravity.

(a) Relative acceleration.

In the given case of Earth-Object system, the force on object due to earth will be equal in magnitude and opposite in direction to the force on earth due the object. The acceleration of the object due to gravity of earth is given by

$g_1=\frac{G{M}_E}{r^2}$

And the acceleration of earth due to gravity of object is given by

$g_2=\frac{Gm}{r^2}$

And therefore, the relative acceleration would be the addition of the two accelerations

$\begin{array}{rcl}{g}_1+g_2& =& \frac{G{M}_E}{r^2}+\frac{Gm}{r^2}\\ a_{rel}& =& \frac{G}{r^2}\left(M_E+m\right)\\ & =& \frac{G{M}_E}{r^2}\left(1+\frac{m}{M_E}\right)\\ & & \end{array}$

Thus, the relation of the relative acceleration with mass of the object is found to be

$a_{rel}=\frac{G{M}_E}{r^2}\left(1+\frac{m}{M_E}\right)$