# Question #de828

Oct 1, 2015

The distance between the earth and the body must be 259 358 400 m for the gravitational attraction of the sun and the earth on the body being balanced.

#### Explanation:

The gravitational attraction between two bodies is calculated as follows:

$F = \frac{G \cdot {m}_{1} \cdot {m}_{2}}{d} ^ 2$

with $G$ the universal gravity constant, ${m}_{1}$ and ${m}_{2}$ the masses of the two bodies and $d$ the distance between the two bodies.

If we place a body on a straight line between the earth and the sun, the resulting gravitational attractions will be:

${F}_{s b} = \frac{G \cdot {m}_{s} \cdot {m}_{b}}{{\left({d}_{s b}\right)}^{2}}$

${F}_{e b} = \frac{G \cdot {m}_{e} \cdot {m}_{b}}{{\left({d}_{e b}\right)}^{2}}$

We are considering the situation when ${F}_{s b} = {F}_{e b}$:

$\frac{\cancel{G} \cdot {m}_{s} \cdot \cancel{{m}_{b}}}{{\left({d}_{s b}\right)}^{2}} = \frac{\cancel{G} \cdot {m}_{e} \cdot \cancel{{m}_{b}}}{{\left({d}_{e b}\right)}^{2}}$

$\rightarrow \frac{{m}_{s}}{{\left({d}_{s b}\right)}^{2}} = \frac{{m}_{e}}{{\left({d}_{e b}\right)}^{2}}$

$\rightarrow \frac{{m}_{s}}{{m}_{e}} = \frac{{\left({d}_{s b}\right)}^{2}}{{\left({d}_{e b}\right)}^{2}} = {\left(\frac{{d}_{s b}}{{d}_{e b}}\right)}^{2}$

$\rightarrow \sqrt{\frac{{m}_{s}}{{m}_{e}}} = \frac{{d}_{s b}}{{d}_{e b}}$

Knowing that ${m}_{e} = 6 \cdot {10}^{24}$kg and ${m}_{s} = 2 \cdot {10}^{30}$kg, and that ${d}_{s b} + {d}_{e b} = 1.5 \cdot {10}^{11}$m we have to solve the following equation:

$\sqrt{\frac{2 \cdot {10}^{30}}{6 \cdot {10}^{24}}} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{{d}_{e b}}$

$\rightarrow {d}_{e b} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{\sqrt{\frac{2 \cdot {10}^{30}}{6 \cdot {10}^{24}}}} = \frac{1.5 \cdot {10}^{11} - {d}_{e b}}{\sqrt{\frac{1}{3} \cdot {10}^{6}}} = \frac{\left(1.5 \cdot {10}^{11} - {d}_{e b}\right) \cdot \sqrt{3}}{{10}^{3}}$

$\rightarrow {d}_{e b} \cdot {10}^{3} + \sqrt{3} \cdot {d}_{e b} = 1.5 \sqrt{3} \cdot {10}^{11}$

$\rightarrow {d}_{e b} \left({10}^{3} + \sqrt{3}\right) = \sqrt{6.75} \cdot {10}^{11}$

$\rightarrow {d}_{e b} = \frac{\sqrt{6.75} \cdot {10}^{11}}{{10}^{3} + \sqrt{3}} \approx 259358400$m