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If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how...

Question:

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how far from the Earth's centre is it when its speed is 1/9 of the escape speed?

Answer and Explanation:

The distance from the Earth's center at the moment when the speed is already 1/9 the escape speed is {eq}5.17 \times10^{8} m {/eq} away

Solution:

The escape speed needed to escape Earth's pull is

{eq}v=\sqrt{\frac{2GM}{r}} {/eq}

{eq}=\sqrt{\frac{2(6.67\times 10^{-11} \frac{m^{3}}{kg s^{2}})(5.97\times10^{24} kg)}{(6.38\times10^{6} m)}} {/eq}

={eq}11,172.61 \frac{m}{s} {/eq}

1/9 the escape speed is

{eq}=1241.4 \frac{m}{s} {/eq}

Rearranging the formula for escape speed

{eq}r=\frac{2GM}{v^{2}} {/eq}

={eq}\frac{2(6.67\times10^{-11} \frac{m^{3}}{kg s^{2}})(5.97\times10^{24})}{(1241.4 \frac{m}{s})^{2}} {/eq}

={eq}5.17\times10^{8} m {/eq}


Learn more about this topic:

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Circular Velocity & Escape Velocity

from Basics of Astronomy

Chapter 25 / Lesson 5
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