Video Transcript
A spacecraft that has been launched
into space is moving along a circular path around Earth. The radius of the orbit is 7,000
kilometers, and the spacecraft has a mass of 2,200 kilograms. What is the kinetic energy of the
spacecraft? Use a value of 5.97 times 10 to the
24th kilograms for the mass of Earth and 6.67 times 10 to the negative 11th cubic
meters per kilogram second squared for the universal gravitational constant. Give your answer to three
significant figures.
Okay, so here we have the Earth,
and we’re told that there is a spacecraft, we’ll say that craft is right here, in
circular orbit around it. We’re told that the radius of the
orbit, which is the distance from the center of the spacecraft to the center of the
Earth, is 7,000 kilometers. And we’ll call that distance
𝑟. We’re also told that this orbiting
spacecraft has a mass of 2,200 kilograms. And we’ll call that mass lowercase
𝑚. We want to know what is the kinetic
energy of the spacecraft. And we’re given values to use for
the mass of the Earth and the universal gravitational constant.
Now, if we recall that, in general,
an object’s kinetic energy is given by one-half its mass times its speed squared, we
might be confused about how to solve for the spacecraft’s kinetic energy because we
don’t know its speed. However, we do know that this
spacecraft is moving in a circular orbit, and any object moving in a circular arc is
subject to what is called a centripetal or center-seeking force. We can call it 𝐹 sub 𝑐.
This centripetal force acting on an
object as it moves in a circle of radius 𝑟 is given by the object’s mass times its
speed squared divided by that radius. Now, any center-seeking force needs
to have some physical mechanism that causes it. In this case, our spacecraft is
moving in a circle around the Earth thanks to the gravitational attraction between
the two masses. In general, the force of gravity
between two masses, we’ll call them uppercase and lowercase 𝑚, is equal to the
product of those masses times the universal gravitational constant divided by the
distance between their centers of mass squared.
What we’re saying is that in the
case of our satellite, the gravitational force experienced by it is equal to the
center-seeking force that makes it move in a circle. In other words, the centripetal
force 𝑚𝑣 squared over 𝑟 is equal to the gravitational force, where we’re saying
that lowercase 𝑚 is the mass of our satellite and uppercase 𝑀 is the mass of the
body being orbited, in this case, Earth.
In this equation, the mass of our
satellite cancels out, as does one factor of the radius 𝑟. And so we find this expression: 𝑣
squared, the speed of our satellite squared as it orbits the Earth, is equal to 𝐺
capital 𝑀 over 𝑟. And this now solves our problem of
not knowing the speed of our satellite in order to calculate its kinetic energy. Now that we know its speed squared
and can express that in terms of these factors, we can say that the kinetic energy
of our orbiting spacecraft is equal to one-half its mass times 𝐺, the universal
gravitational constant, multiplied by the mass of, in this case, the Earth, divided
by the distance between our satellite and the center of the Earth.
And we can now notice that we’re
given the masses of these two bodies as well as the universal gravitational constant
and this distance 𝑟. All that remains, then, is for us
to substitute in these values and then calculate the kinetic energy. So here we have our satellite mass,
2,200 kilograms; our value for the universal gravitational constant; our mass of the
Earth, 5.97 times 10 to the 24th kilograms; and 𝑟, 7,000 kilometers.
There’s just one change we’ll want
to make before calculating kinetic energy. And it comes down to what is right
now a disagreement in units. Notice that in our value for the
universal gravitational constant, we have units of distance given in meters, whereas
our radius is currently in kilometers. If we recall, though, that 1,000
meters is equal to one kilometer, then that tells us that 7,000 kilometers is equal
to 7,000 with three zeros added to the end meters. That is seven million meters. Now, the units all throughout our
expression do agree, and we can go ahead and compute the kinetic energy of the
spacecraft.
To three significant figures, we
find a result of 6.26 times 10 to the 10th joules. And if we recall that 10 to the
ninth, or a billion joules, is equal to one gigajoule, then we can express our
answer as 62.6 gigajoules. This is the spacecraft’s kinetic
energy, which, notice, is constant all throughout its circular orbit.
