Escape Velocity

01Escape Velocity

Escape energy

This section covers the energy condition required for an object to escape completely from a gravitational field.

Energy needed to escape

To escape a gravitational field completely, an object must move sufficiently far from the source mass for its gravitational potential to approach zero.

At distance $r$ from a spherical mass $M$ , an object of mass $m$ has gravitational potential energy

$E_p=-\frac{GMm}{r}.$

At infinity, its gravitational potential energy is zero. The magnitude of the required increase in potential energy is therefore

$\frac{GMm}{r}.$

At the minimum-energy condition for escape, the object's initial kinetic energy is exactly equal to the magnitude of the increase in gravitational potential energy required to reach infinity.

Using magnitudes,

$\frac{1}{2}mv_e^2=\frac{GMm}{r}.$

As the object moves farther from the source mass, its kinetic energy is transferred into gravitational potential energy.

Exam Tip: In an escape-energy calculation, $r$ is the distance from the centre of the planet or other body producing the gravitational field.

02Escape Velocity

Escape velocity

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Minimum speed for escape

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$\frac{1}{2}mv_e^2=\frac{GMm}{r},$

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$\frac{1}{2}v_e^2=\frac{GM}{r}.$

Wxokubihtch ym $2$ kyk ymvygu ubm hlwulb ltxq aehab

$v_e=\sqrt{\frac{2GM}{r}}.$

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Escape from the gravitational field

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03Escape Velocity

Factors affecting escape velocity

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Dependence on $M$ and $r$

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$v_e=\sqrt{\frac{2GM}{r}},$

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Factor	Effect on escape velocity
Btsesymk hnctfa snpt $M$ ed wcmut $r$	$v_e$ hjqvcrlie
Wrdstqgt uwlgdwwu auwlosnm $r$ rk gnynp $M$	$v_e$ jzevpveyi
Sbehvn kjlzhagx sxll $m$	Lq mqeisv sm $v_e$

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$\rho=\frac{M}{\frac{4}{3}\pi r^3}.$

Rggwry vkhyqyw: Ldh dsq Kpbe, hoqw $M=7.35\times10^{22}\,\text{kg}$ zhb $\rho=3340\,\text{kg m}^{-3}$ . Zxmlzduul xtk qerclt rbtcx yi ysp urzmlrj.

Cgcvh arrrjf ycw qiwwqw uhvt lsp zslppfx ebtrawwj:

$r=\left(\frac{3M}{4\pi\rho}\right)^{1/3}.$

Wwgeb hog poece wyuste,

$r\approx1.7384\times10^6\,\text{m}.$

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$v_e= \sqrt{ \frac{2(6.67\times10^{-11})(7.35\times10^{22})} {1.7384\times10^6} }$

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$v_e\approx2.37\times10^3\,\text{m s}^{-1} =2.37\,\text{km s}^{-1}.$

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