Video Transcript
In this video, we’re going to learn
about escape velocity. We’ll see what this term means. And we’ll also find out how to
calculate it in different scenarios.
To start out, imagine that you’re
in a model rocket competition, where in this competition, a prize is awarded to the
person whose rocket travels highest. In order to guarantee victory, you
want your rocket not only to travel higher than the others, but never to return to
Earth. You want it to completely escape
Earth’s gravitational pull.
In order to understand what will be
needed for your rocket to do this, it will be helpful to learn about escape
velocity. This term “escape velocity” comes
out of a context of considering how to escape a planet’s gravitational field. Imagine that, in the entire
universe, there’s just one planet in existence. Everything outside of that planet
is just empty space.
We know that a planet having mass
will create a gravitational field around itself, which will tend to draw other
massive objects in towards itself. If we have some other massive
object then on the surface of the planet, the tendency of that object is to stay
there on the surface. Energetically, that’s what the
object wants to do.
There’s a helpful way of showing
this graphically. Imagine that we zoomed way out from
our planet so that it looked like this. If we were to make a graph of the
gravitational potential energy 𝑢 created by the planet, as a function of the
distance 𝑟 from the planet’s surface, that curve would look something like
this. Following the functional form GPE
is equal to negative the universal gravitational constant times the mass of our
planet times the mass of the object it’s attracting, all divided by the distance
between their centers.
We can notice that this curve, as
it moves farther and farther away from the planet, approaches zero but never reaches
it. We can think of this curve almost
as though it’s a hill, something that starts at the surface of the planet we’re
talking of and climbs up and up and up, requiring energy to move further away from
the planet. Let’s say that some object which
started out on the surface of the planet was energetic enough to climb up this curve
farther and farther until eventually, as it continues to move out, it effectively
escapes the gravitational pull of the planet.
In order to climb out of this
gravitational well or to climb up to the top of this gravitational hill, our object
will need to start out on the surface of the planet with a certain minimum
speed. That speed, which along with the
mass of the object gives it a kinetic energy, is what is paid out over its ascent up
this hill, until finally it escapes the gravity of the planet. This object by the way doesn’t need
to be infinitely far away from the planet in order to have achieved an escape
velocity. It’s more a question of the energy
gap to be covered which is not infinite but finite.
Escape velocity then, which we can
call 𝑣 sub 𝑒, is all about energy. It’s about an object’s kinetic
energy being enough to meet or overcome a planet’s gravitational potential
energy. We can say that an object’s escape
velocity is found by balancing the object’s kinetic energy with a planet’s
gravitational potential energy. Since escape velocity, 𝑣 sub 𝑒,
is actually a speed rather than a velocity, we’ll equate the kinetic energy of the
escaping object with the magnitude of the planet’s gravitational potential
energy.
In this mathematical relationship,
there’s something very interesting and perhaps counterintuitive. We see that the mass of our object,
which we want to escape the gravitational pull of the planet, appears on both sides
of the equation. Meaning that, mathematically, the
mass cancels out so that, apparently, an object’s escape velocity has nothing to do
with its own mass.
Canceling that mass out and
dropping the absolute value bars, we can rearrange now to solve for 𝑣 sub 𝑒, the
escape velocity. We find that it’s equal to the
square root of two times the universal gravitational constant times the mass of the
spherical body it’s trying to escape from divided by the radius of that body. Notice that the right side of this
equation has nothing in it about the object that’s escaping but is only about the
object being escaped from. And in particular, for a
spherically symmetric body, it only depends on that body’s mass and its radius.
It turns out when we enter in the
numbers for planet Earth that the escape velocity for an object to escape Earth’s
gravitational field is about 11 kilometers per second. If an object can achieve that
speed, it will be able to leave Earth’s gravity and escape the planet
permanently. We can calculate escape velocity
though for any spherically symmetric object.
Say that we have a tennis ball. The mass of the tennis ball is
about 0.058 kilograms. And it has a radius of a little
over three centimeters. If we enter these values in SI base
units into our equation for escape velocity along with the universal gravitational
constant, we find that, for an object to escape from the surface of a tennis ball,
it needs to have a minimum speed of about 1.5 times 10 to the negative fifth meters
per second, which is roughly an object moving at one twentieth of a meter every
hour.
This just shows that escape
velocity can be calculated for any spherically symmetric object. But of course, it was developed in
a context of planetary scales. Let’s get some more practice
calculating escape velocity using another example.
What is the speed needed to escape
from the Earth–Moon system from a point on the surface of Earth? Assume there are no other bodies
involved. And do not account for the fact
that Earth and the Moon are moving in their orbits. Use a value of 385.0 times 10 to
the third kilometers for the distance between the centers of the Moon and Earth,
5.97 times 10 to the 24th kilograms for Earth’s mass, and 7.35 times 10 to the 22
kilograms for the Moon’s mass.
We’re asked to solve for the speed
needed to escape the Earth–Moon system. That’s an escape velocity which
we’ll call 𝑣 sub 𝑒. And we can start on our solution by
drawing a sketch of this situation. Considering this system of the
Earth and the Moon which orbits around it, when we consider an object somewhere on
Earth’s surface escaping the gravitational pull of this system, we’d want to
position it on Earth so that we minimize the escape speed or escape velocity needed
to do this.
The position on Earth’s surface
then that we pick for our object would be directly opposite the position of the
Moon. That way, our mass is already as
far as possible from the center of mass of the Earth–Moon system. So if our mass was to blast off
perpendicularly from Earth’s surface, in order to escape this system, what escape
velocity would it need?
We recognize this as an energy
balance situation. The kinetic energy of our escaping
object must be at least equal to the gravitational potential energy created by the
Earth and the Moon on that object. Recalling that kinetic energy is
equal to an object’s mass times its speed squared divided by two. And that the magnitude of the
gravitational potential energy between two masses, capital 𝑀 and lowercase 𝑚, is
equal to their product times the universal gravitational constant divided by the
distance between their centers of mass. From an energy balance perspective,
we can write that the kinetic energy of our escaping mass is equal to the
gravitational potential energy between that mass and the Earth plus the
gravitational potential energy between that mass and the Moon.
Looking over this expression, we
see that our escaping object’s mass appears in every term. And therefore, it cancels out. If we then multiply both sides of
the equation by two and take the square root of both sides, we find that the escape
velocity of our object is equal to the square root of two times big 𝐺, all
multiplied by the mass of the Earth over its radius plus the mass of the moon over
the radius of the Earth plus the distance between the center of the Moon and the
Earth. Big 𝐺, the universal gravitational
constant, we set to be exactly equal to 6.67 times 10 to the negative 11th cubic
meters per kilogram second squared.
Looking over this equation, we see
that we know all of the values in it except for 𝑟 sub 𝐸, the radius of the
Earth. That’s a value we can look up. And when we do, we find it’s equal
to approximately 6.37 times 10 to the sixth meters. We’re now ready to plug in and
solve for 𝑣 sub 𝑒, the escape velocity of our object.
When we do plug in, we’re careful to
convert our distance between the centers of the Moon and the Earth into units of
meters so it agrees with the units of the rest of this expression. To three significant figures, 𝑣
sub 𝑒 is 11.2 times 10 to the third meters per second. That’s how fast an object would
need to be moving off of Earth’s surface in order to escape the Earth–Moon
system.
Let’s summarize what we’ve learned
so far about escape velocity. We’ve seen that escape velocity,
often symbolized 𝑣 sub 𝑒, is the speed needed to leave an object’s gravitational
field. We’ve also seen that escape
velocity is found using energy balancing. We say that when the initial
kinetic energy of our object that wants to escape is equal to negative the
gravitational potential energy of the object that draws it, then that escaping
object’s speed will be equal to its escape velocity.
For a spherically symmetric object,
this relationship implies that the escape velocity is equal to the square root of
two times the universal gravitational constant times the mass of the spherical body
being escaped from, all divided by its radius. And finally, we’ve seen that an
object’s escape velocity is independent of its mass. And it only varies with the planet
it’s escaping from’s mass and radius.
