Escape Velocity – Part 1

As the reader probably knows, any planet or moon has an Escape Velocity. A mass at the surface travelling at this speed or greater will go infinitely far – gravity will never be able to pull it back.

This post is about escape velocity, including the formula for computing it.

To compute escape velocity, we use energy concepts, specifically kinetic and potential energy. As we learn in high school physics, if we have a mass \( m \), it has a weight at the Earth’s surface equal to \( mg \), where \( g \) is the gravitational acceleration constant for Earth ( \( \text{9.8 meters / sec}^2\) in MKS units). If we lift the mass a distance \( h \), it gains potential energy in the amount \( mg \times h \) . If we then drop it, it begins loosing potential energy, but the decrease in potential energy is converted to kinetic energy \( 1/2 \, m v^2 \). When it reaches the starting point, the potential energy is zero, and all of the energy is kinetic, so that

\[ \frac{1}{2}  m v_0^2 = mgh \tag{1} \]

Solving for \( v_0 \),

\[ v_0 = \sqrt{2gh} \tag{2} \]

The exchange also works in reverse: if we toss it upwards at speed \( v_0 \), it will reach height \( h \), at which point  the kinetic energy is zero.

For escape velocity, we do a similar calculation, but as we move away from the Earth, \( g \) is not a constant. As the law of gravity tells us,  \( g \) decreases as the inverse square of distance from Earth’s center, so \( g \) varies as

\[ g = g_0  \left( \frac{r}{R_0} \right)^2 \tag{3}\]

Where \( g_0 \) is 9.8, \( r \) is distance from Earth’s center, and \( R_0 \) is the Earth’s radius. This is graphed below.

Acceleration value g as a function of Earth radii

Acceleration value g as a function of Earth radii

We want to find the potential energy gain in moving a unit mass from \( R_0 \) to infinity. If we know that, we can set it equal to \( 1 / 2\, m v_{escape}^2 \), and solve for \( v_{escape} \).

The graph below shows how we might at least approximate the potential energy required .

Approximating the potential energy

Approximating the potential energy

If we divide the distance into equal intervals of width \( \Delta r \), we could assume that over each interval, \( g \) is constant. The potential energy gain in each interval is then \( g \times \Delta r \), where \( g \) is the height of the rectangle. The PE gain is thus the area of the rectangle, and the total PE is approximately the sum of all the rectangle areas.

Obviously, the approximation gets better as the rectangles are made narrower. If the rectangles are made infinitely narrow, the total potential energy just becomes the area under the curve, as shown in the graph below.

Exact Potential Energy

Exact Potential Energy

A bit of calculus can show that the shaded area  is precisely \( g_0 R_0 \). Therefore,

\[ g_0 R_0 =\frac{1}{2} v_{escape}^2 \]


\[ v_{escape} = \sqrt{2 g_0 R_0} = \sqrt{ 2 \times 9.8 \times 6.37  \times 10^6} = 11.17  \text{km/sec} \tag{4}\]

This equates to 40,226 km/hour or 24,995 miles per hour.

Note that we have not assumed that the launch direction is straight up. It’s a matter of energy; at escape velocity, the energy is sufficient to escape the gravity, no matter the direction. Isaac Newton showed that the path the mass will follow is a parabola, with the center of the planet at the focus of the parabola (a straight line is a special case of a parabola).

Another version of the formula

Obviously, formula (4) is specific to the Earth. We usually see the formula derived by starting with the gravity equation, where the force on a unit mass is:

\[ \frac{GM}{r^2} \]

\( G \) is the universal gravitational constant, \( M \) is the mass of the attracting body (e.g., the Earth), and \( r \) is distance from the body’s center. When we apply the same process as above to this equation, we get the general formula:

\[ v_{escape} = \sqrt{\frac{2GM}{R_0}} \tag{5} \]

Using the mass and radius for Earth, this  produces the same value as formula (4).

In the last graph, the curve extends to infinity on the right, and it is not obvious that the total area will be a finite amount. As we have seen, the area is in fact finite. Suppose, however, that gravity decreased as \( 1/r \), instead of \( 1 / r^2 \). In that case, the graph will look similar, but  the area under the curve (and thus the required energy) turns out to be infinite. Thus, in a \( 1/r \) universe, it would be impossible to have enough velocity and kinetic energy to fully escape from  gravity.

In the next post, I will go into some implications of the escape velocity formula.


Feynman Lectures on Physics, Vol 1 by Richard Feynman



4 Responses to “Escape Velocity – Part 1”

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  2. David Rosen Says:

    All of this assumes that for the object to escape, its kinetic energy must not reach zero before it reaches infinity. There is one point about this which is a bit confusing to me:

    Let’s say the object is launched below escape velocity but fast enough to establish an elliptical orbit. We seem to be saying that it will reach zero kinetic energy at the furthest point of the orbit (apoapsis). However, it seems that its velocity would not be zero at that point. Only the component of the velocity directly away from the earth would be zero. But would still retain tangential velocity and so would still retain kinetic energy. Therefore kinetic energy would not in fact be zero.
    What is the flaw in this reasoning?
    Thanks for your help!

  3. curiousCharacter Says:

    Good question, David Rosen, and I see what you are thinking.
    First consider the potential energy PE. It is zero at the Earth’s surface. At infinity, it is some positive amount given by the formula. At a finite distance, it has some intermediate value.
    Say we launch with an initial kinetic energy KE1, and KE1 is is not enough to escape gravity. Conservation of energy requires that at all subsequent times we must have KE + PE = KE1. If the launch direction is directly away from the Earth’s center, it will reach a maximum distance when all the kinetic energy has been used up, and velocity is indeed zero, before it starts falling back. This is actually an infinitely flat elliptical orbit (eccentricity = 1).
    If the launch direction is not directly outward, Newton’s mathematics show that it will follow an ordinary elliptical orbit, and will reach apoapsis before the velocity is zero. That is, for an ordinary orbit, the initial velocity is never entirely “used up”.
    I hope this helps.

  4. Hayden White Says:

    The principle of launching a satellite is to impart a sufficient horizontal initial speed(escape velocity) so that it should go out side the earth are move around the earth in an orbit. The velocity(escape velocity) should be 11.2 km/sec.

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