Video Transcript
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.
