### Video Transcript

Find the mass of the Sun. Use a value of 152 times 10 to the
sixth kilometers for the average distance between the centers of the Earth and the
Sun.

We can make progress solving for
the mass of the Sun by considering the Earth as it rotates about the Sun. Throughout this rotation, thereβs a
gravitational force acting on the Earth and pulling it into the center of the
Sun.

Thanks to Newtonβs law of universal
gravitation, we can say that the gravitational force between any two masses, π one
and π two, is equal to their product times the universal gravitational constant
divided by the distance between them squared. In the case of the Earth in its
orbit, this gravitational force causes the Earth to move in a circle. In other words, itβs a centripetal
force.

We can further recall then that the
centripetal force acting on an object is equal to the mass of that object times its
linear speed squared divided by the radius of the circle that the object moves
in. This means we can equate the
gravitational force between the Sun and the Earth and the centripetal force acting
on the Earth.

In this expression, the mass of the
Earth, π sub π, cancels out since itβs common to both sides, as does a factor of
the radius π in the denominator. If we then rearrange this
expression to solve for π sub π , the mass of the Sun, we find that itβs equal to
π times π£ squared over πΊ, where π£ is the linear speed of the Earth as it moves
around its circular orbit.

Recalling that, for an average
speed π£, that speed is equal to distance traveled divided by the time it takes to
travel that distance, we can say that the speed π£ of the Earth in its orbit around
the Sun is the distance it travels one circumference of that orbit, two ππ,
divided by the time it takes to travel that orbit, the period π. When we plug this expression for π£
into our equation for the mass of the Sun, we see that π sub π is equal to four π
squared π cubed over π squared, the period of the Earth in its orbit, times the
universal gravitational constant, πΊ.

In the problem statement, weβre
told what π is. Itβs 152 times 10 to the sixth
kilometers. We can look up the universal
gravitational constant, finding a value of 6.67 times 10 to the negative 11th cubic
meters per kilogram second squared. We can enter in these values for π
and πΊ into our equation for the mass of the Sun. And as we do, we convert the units
of π from kilometers to meters.

As a last step, weβll want to solve
for the period π of the Earthβs rotation around the Sun. How long does that rotation
take? We know that rotation is one year,
which is approximately 365 days, where each day has 24 hours and each hour has 3600
seconds. So 365 times 24 times 3600 is about
the number of seconds in a year. To three significant figures, we
find that π sub π is 2.09 times 10 to the 30th kilograms. Thatβs the mass of the Sun
calculated based on its gravitational attraction to the Earth.