Calculate the magnitude of the average linear momentum of the Earth in its orbit. Use a value of 5.97 times 10 to the 24th kilograms for the mass of the Earth and use a value of 1.50 times 10 to the 11th metres for the average orbital radius of the Earth.
As we work this problem, we’ll assume that the Earth moves in a perfect circle in its orbit around the Sun. Let’s begin by making a sketch of this scenario. Here, we have the Sun, the centre, and the Earth, the small blue dot, that moves in an orbit, a circular orbit, around that central Sun.
In our problem statement, we’re told what the mass of the Earth is: 5.97 times 10 to the 24th kilograms. And we’re also told that on average, the Earth is at distance of 1.50 times 10 to the 11th metres away from the centre of the Sun. We’re looking to solve for the momentum of the Earth in its orbit and we’ll represent that momentum using the letter lowercase 𝑝.
Let’s recall the equation that tells us what the momentum of an object is. The momentum of an object is defined as the mass of that object times its velocity. So in our case, if we know the mass and the speed of the Earth, then we’ll be able to solve for our average linear momentum. The mass of the Earth is given to us. We’ll call that 𝑚 sub 𝑒. Knowing the mass and knowing that the momentum that we seek is equal to the mass times the speed of the Earth, we now want to figure out what is that speed of the Earth in its orbit.
We’re told the average radius of the Earth’s orbit as it moves around the sun. We can call that value 𝑟. And it equals 1.50 times 10 to the 11th metres. Now recall the simple equation for linear speed. That the speed of an object is equal to the distance it travels divided by the time it takes to travel that distance. This equation tells us that if we can solve for the distance that the Earth travels in its orbit and know how long that takes, then we can solve for 𝑉, the speed of the Earth as it moves to its orbit.
So what is 𝑑, the distance the Earth travels in its orbit, and 𝑡, the time it takes to travel that far? First, 𝑑, that will be the circumference of the circle defined by the radius given in the problem statement. When the Earth travels one full revolution around that circle, it’s completed one orbit. And that distance is equal to the diameter of the orbit multiplied by 𝜋. The diameter of that circle is equal to twice the radius. So we can write out the total distance travelled by the Earth in one orbit. It’s equal to two times the radius times 𝜋.
Now for the time it takes for the Earth to make one complete revolution around the Sun, we know that that equals roughly 365 days. But looking at the units of the speed, currently we have metres per day, but we would like to have the units in metres per second. So let’s convert our time from days to seconds.
How many seconds are in a year? That will be equal to the number of days times the number of hours in a day, 24, times the number of seconds in an hour, 3600. This will give us a unit in our denominator of seconds. Entering these values into our calculator, we find that the linear speed of the Earth in its orbit is equal to 3.00 times 10 to the fourth metres per second.
We can now use the speed in our equation for momentum to solve for the average linear momentum of the Earth in its orbit. Plugging in our values for 𝑚 sub 𝑒 and 𝑉, when we multiply these numbers together, we find a total average linear momentum of the Earth of 1.78 times 10 to the 29th kilograms metres per second. This is the average linear momentum of the Earth in its orbit.