Escape Speed

$$v_\text{escape} = \sqrt{\frac{2GM\phantom{)}}{R}}\quad[\text{Eq. (6.7)}]\quad\text{or}\quad v_\text{escape} = 11.2~\text{km}/\text{s}\times\sqrt{\frac{M_\text{Earth}}{R_\text{Earth}}}$$

Note: If you feel uncomfortable with working with variables, fractions, percentages, or scientific notation, you should review the following tutorials before working through this tutorial:

Sections

Introduction

The escape speed equation is the result of asking the question: How fast must an object travel straight up in order to escape from a body of mass \(M\) that has a radius of \(R\) (note that we are ignoring any air resistance)? To answer the question it first gets converted into a conservation of energy equation: What must the kinetic energy \(\frac{1}{2}mv^2\) of the object be in order to overcome the change in potential energy due to gravity in going straight up from the surface of the body to infinitely far away, or \(\frac{1}{2}mv_\text{escape}^2 = \text{change in gravitational potential energy}\).

Advanced: For the curious, the change in gravitational potential energy that goes on the right-hand side of the expression derives directly from the force equation that is Newton’s universal law of gravitation [Equation (5.7)]: \(\text{change in gravitational potential energy} = GMm/R\). The 2 on the right-hand side of Equation (6.7) comes from the \(\frac{1}{2}\) in the kinetic energy expression the little \(m\)s cancel, and the square root comes from solving for \(v_\text{escape}\). BTW, whenever you see Big G (\(G\)) you know gravity is involved.

The alternate form of the escape speed equation comes from using the “trick” used many times in the textbook: divide the general equation by the same equation using specific values of a well-known object or system. The obvious “well-known” object that is used here is Earth. Substituting the mass of Earth (\(\text{M}_\text{Earth}\)), the radius of Earth (\(\text{R}_\text{Earth}\)), and the value of Big G, tells us that the escape speed from Earth is \(11.2~\text{km}/\text{s}=\sqrt{2G\text{M}_\text{Earth}/\text{R}_\text{Earth}}\). Dividing the general equation [(Equation (6.7)] by this last equation produces the alternate form of the escape speed, where the mass and radius MUST be expressed in terms of the mass and radius of Earth (the Earth subscripts are meant to remind you of this requirement in using the alternate form).

Examples

  1. Venus is very similar in size to Earth. Its mass is \(81\%\) of Earth’s mass and its radius is \(95\%\) of Earth’s radius. What is the escape speed from the surface of Venus? $$v_\text{escape} = 11.2~\text{km}/\text{s}\times \sqrt{\frac{0.81}{0.95}} = 11.2~\text{km}/\text{s}\times\sqrt{0.85} = 11.2~\text{km}/\text{s}\times 0.92 = 10.3~\text{km}/\text{s}.$$
  2. The Sun’s mass is \(\text{333,000}\) times the mass of Earth and its radius \(109\) times Earth’s radius. What is the Sun’s escape speed? $$v_\text{escape} = 11.2~\text{km}/\text{s}\times\sqrt{\frac{\text{333,000}}{109}} = 11.2~\text{km}/\text{s}\times \sqrt{3055} = 11.2~\text{km}/\text{s}\times55.3=619~\text{km}/\text{s}.$$
  3. Ceres is located in the asteroid belt and is classified as both an asteroid and a dwarf planet. It’s mass is \(0.000\,16\) times the mass of Earth, and its radius is only \(0.075\) times Earth’s radius. What is its escape speed? $$v_\text{escape}=11.2~\text{km}/\text{s}\times\sqrt{\frac{0.000\,16}{0.075}} = 11.2~\text{km}/\text{s}\times\sqrt{0.0021}=11.2~\text{km}/\text{s}\times0.046=520~\text{m}/\text{s}$$.
  4. Mimas, a small moon of Saturn, has a mass estimated to be \(6.3 \times 10^{-6}\) times the mass of Earth, and a radius of \(198~\text{km}\) (the radius of Earth is \(6378~\text{km}\)). Calculate the escape speed of Mimas. $$v_\text{escape} = 11.2~\text{km}/\text{s}\times \sqrt{\frac{6.3\times 10^{-6}}{198/6378}} = 11.2~\text{km}/\text{s}\times \sqrt{\frac{6.3\times 10^{-6}}{0.031}} =160~\text{m}/\text{s}.$$

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers are available below.)

  1. Suppose that Earth’s mass magically increased by a factor of 9 without any change in its radius. Would its escape speed increase or decrease, and by what factor?
  2. Suppose that Earth’s mass increases by a factor of 2 while its radius decreases by a factor of 8. What would the new value of Earth’s escape speed be?
  3. Jupiter has a mass of \(318~\text{M}_\text{Earth}\) and a radius of \(11.2~\text{R}_\text{Earth}\). What is the escape speed from the “surface” of Jupiter. (Actually Jupiter is a gas giant and doesn’t have a solid surface.)
  4. You land on a planet with twice Earth’s radius and eight times Earth’s mass. Determine the planet’s escape speed. (You shouldn’t need to use a calculator for this one.)
  5. The mass of Mimas, one of Saturn’s moons, is \(0.000\,0063~\text{M}_\text{s}\) and its radius is \(198~\text{km}.\) The radius of Earth is \(6378~\text{km}\). What is the escape speed of Mimas?
  6. Bennu is a near-Earth asteroid with an estimated mass of \(7.3\times 10^{10}~\text{kg}\) and an average radius of \(245~\text{m}\). Earth’s mass and radius are \(5.97\times 10^{24}~\text{kg}\) and \(6378~\text{km}\), respectively. (a) What is the escape speed from the surface of Bennu? (b) A major-league fastball is about \(45~\text{m}/\text{s}\); the world record top running speed is \(12.4~\text{m}/\text{s}\); a human typically walks at roughly \(2~\text{m}/\text{s}\). Which one of these speeds is the escape speed of Bennu closest to, and how many times faster or slower is Bennu’s escape speed? (c) Could someone jump off of Bennu’s surface? (A space mission has traveled to Bennu to retrieve some of its surface material to return to Earth.)
  7. Sirius B is a star that has roughly the radius of Earth, but has a mass that is close to the mass of the Sun (about 333,000 times the mass of Earth). What is the escape speed of Sirius B?

Answers

  1. The escape speed would increase by a factor of 3.
  2. \(5.6\text{ km}/\text{s}\)
  3. \(59.7~\text{km}/\text{s}\)
  4. \(22.4~\text{km}/\text{s}\)
  5. \(0.16~\text{km}/\text{s}\)
  6. (a) \(0.0002~\text{km}/\text{s} = 0.2~\text{m}/\text{s}\). (b) Bennu’s escape speed is 10 times slower than typical walking speed. (c) Yes, and easily!
  7. \(6500\text{ km}/\text{s}\) (about 4040 miles per hour)
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