Video: Calculating the Mass of the Sun

Find the mass of the Sun. Use a value of 152 Γ— 10⁢ km for the average distance between the centers of Earth and the Sun.

03:07

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, there’s a gravitational force acting on the Earth and pulling it into the center of the Sun.

Thanks to Newton’s law of universal gravitation, we can say that the gravitational force between any two masses, π‘š one and π‘š 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, it’s 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‘š sub 𝑒, cancels out since it’s common to both sides, as does a factor of the radius π‘Ÿ in the denominator. If we then rearrange this expression to solve for π‘š sub 𝑠, the mass of the Sun, we find that it’s equal to π‘Ÿ times 𝑣 squared over 𝐺, where 𝑣 is the linear speed of the Earth as it moves around its circular orbit.

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

In the problem statement, we’re told what π‘Ÿ is. It’s 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‘Ÿ and 𝐺 into our equation for the mass of the Sun. And as we do, we convert the units of π‘Ÿ from kilometers to meters.

As a last step, we’ll want to solve for the period 𝑇 of the Earth’s 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‘š sub 𝑠 is 2.09 times 10 to the 30th kilograms. That’s the mass of the Sun calculated based on its gravitational attraction to the Earth.

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