Video Transcript
An asteroid has a mass of 4.7 times
10 to the 13th kilograms. The asteroid passes near Earth, and
at its closest approach, the separation of the centers of mass of the asteroid and
Earth is four times the average orbital radius of the Moon. What force does the asteroid exert
on Earth when at its minimum distance from Earth? Use a value of 384400 kilometers
for the average orbital radius of the Moon.
We’ll label this force, we want to
solve for, capital 𝐹 and start off by drawing a sketch of the situation. In this situation, our asteroid
labelled 𝑎 passes by the Earth labelled 𝐸 at a minimum distance of four times the
orbital radius of the moon around the Earth. Given the mass of the asteroid 𝑚
sub 𝑎 and the orbital radius of the moon around the Earth 𝑂𝑅 sub 𝑚, we want to
solve for the gravitational force of attraction between the asteroid and Earth when
they’re closest. To solve for this force, we recall
that the gravitational force of attraction between any two masses, 𝑚 one and 𝑚
two, is equal to their product divided by the square of the distance between their
centers of mass all multiplied by the universal gravitational constant capital
𝐺.
We’ll let that constant 𝐺 be
exactly 6.67 times 10 to the negative 11th cubic meters per kilogram second
squared. When we apply the mathematical
relationship for gravitational force to our scenario, we can say that 𝐹, the force
we wanna solve for, is equal to 𝐺 times the mass of the Earth times the mass of the
asteroid all divided by four times the orbital radius of the Moon quantity
squared. We know the value in the
denominator given to us. And we also know the mass of the
asteroid, as well as the constant 𝐺. All that remains is to solve for
the mass of the Earth. And we can do that by looking up
this value.
A commonly accepted value for the
mass of the Earth is 5.95 times 10 to the 24th kilograms. Knowing that, we’re ready to plug
in and solve for 𝐹. When we do plug these values in,
we’re careful to convert the orbital radius of the Moon into units of meters so that
it’s consistent with the units in the rest of our expression. Speaking of units, let’s take a
second to consider what the final units of this calculation will be. Looking in the numerator of this
overall expression, we see that a factor of kilograms will cancel out. Looking then at the units of meters
that appear in our expression, we see that the meter squared in the denominator will
take away two meter factors in our numerator so that overall we’ll simply have
meters to the first.
The units of second squared will
remain in our overall denominator. So we expect to get final units of
kilograms meters per second squared, which agrees with the units we’d expect for a
force, that is, units of newtons. When we calculate this result, we
find, to two significant figures, that it’s 7.9 times 10 to the ninth newtons. That’s the gravitational force of
attraction between the asteroid and the Earth.
