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If the sun is about $9.3\times10^7\; km $ away from the earth, and its mass is $3.24\times10^5 \times m_e$, where $m_e$ is the mass of the earth, how far from the earth must a body be along a line towards the sun so that the sun’s gravitational pull balances the earth's?

$\begin{array}{1 1} (A) 1.73 \times 10^5 \; km \\(B) 1.63 \times 10^5 \; km \\(C) 1.83 \times 10^5 \; km \\ (D) 1.43 \times 10^5 \; km \end{array}$

1 Answer

Let $m_s$ and $m_e$ be the mass of the sun and earth respectively $\rightarrow m_s =3.24\times10^5 \times m_e$.
Let the body of mass $m$ be at a distance $x$ from the center of earth and distance $d$ be the distance from the center of the sun to center of earth.
If the gravitational forces are balanced $\rightarrow \large\frac{G_m\times m_e}{x^2} $$ = \large\frac{G_m \times m_s}{(d-x)^2}$
Given that $ d = 9.3\times10^7\; km $, we can calculate $x$,
$\Rightarrow \large\frac {x}{9.3\times10^7-x} $$ = \large\sqrt \frac{m_e}{3.24 \times 10^5 me}$
$\Rightarrow 569.2\;x = 9.3\times10^7 - x \rightarrow x \approx \large\frac{9.3\times10^7}{570.2} $$\approx 1.63 \times 10^5 \; km$
answered Aug 23, 2013 by meena.p
edited Mar 23, 2014 by balaji.thirumalai

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