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101P

Expert-verifiedFound in: Page 385

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A spaceship is on a straight-line path between Earth and the Moon. At what ****distance from Earth is the net gravitational force on the spaceship zero?**

Net gravitational force on the spaceship is zero at a distance of $3.44\times {10}^{8}\mathrm{m}$ from the earth.

The distance between the Earth and the Moon is ${\mathrm{R}}_{\mathrm{Earth}}=3.82\times {10}^{8}\mathrm{m}$.

**Let the distance from Earth to the spaceship be r. **

Formula:

$\begin{array}{rcl}{\mathrm{f}}_{\mathrm{E}}& =& \frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\left({\mathrm{R}}_{\mathrm{em}}-\mathrm{r}\right)}^{2}}\\ & =& \frac{{\mathrm{GM}}_{\mathrm{e}}\mathrm{m}}{{\mathrm{r}}^{2}}\end{array}$

where m is the mass of the spaceship.

Solving for r, we obtain

$\begin{array}{rcl}\mathrm{r}& =& \frac{{\mathrm{R}}_{\mathrm{em}}}{\sqrt{\mathrm{M}/{\mathrm{M}}_{\mathrm{e}}}+1}\\ & =& \frac{3.82\times {10}^{8}\mathrm{m}}{\sqrt{\left(7.36\times {10}^{22}\mathrm{kg}/\left(5.98\times {10}^{24}\right)\mathrm{kg}\right)}+1}\\ & =& 3.44\times {10}^{8}\mathrm{m}\end{array}$

The net gravitational force on the spaceship is zero at a distance of $3.44\times {10}^{8}\mathrm{m}$ from the earth.

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