Video Transcript
Geostationary satellites orbit Earth at a height of 35,786 kilometers above the equator. Earth has a mass of 5.97 times 10 to the 24 kilograms and a radius of 6,370 kilometers. What is the ratio of the local acceleration due to gravity at the height of a geostationary satellite to that on the surface of Earth? Give your answer to three decimal places.
A diagram is helpful here to visualize the setup. So here’s Earth and a geostationary satellite in orbit above the equator. We’re told that the satellite orbits at a height of 35,786 kilometers, so we can fill that in on our diagram. And we’re given Earth’s mass of 5.97 times 10 to the 24 kilograms and finally Earth’s radius of 6,370 kilometers. We’re asked to find the ratio of the local acceleration due to gravity at the height of the satellite to that on the surface of Earth.
So if we imagine a person standing on Earth’s surface, we need the ratio of the local acceleration due to gravity experienced by this satellite to this person. So we need to recall the equation for the local acceleration due to gravity. That is 𝑎 is equal to 𝐺𝑚 over 𝑟 squared, where 𝑎 is the acceleration due to gravity, 𝐺 is the universal gravitational constant, 𝑚 is the mass of the large body we’re accelerating towards, and 𝑟 is the distance to the center of mass of the large body, in this case Earth.
Focusing first on the satellite, let’s call the local acceleration due to gravity at the height of a geostationary satellite 𝑎 sub g. This is equal to the universal gravitational constant big 𝐺 times 𝑚, the mass of the Earth, divided by the distance between the center of Earth to the height of a geostationary satellite, we’ll call that 𝑟 sub g squared. Now let’s compare that to the local acceleration due to gravity on the surface of Earth. Let’s call that 𝑎 sub e. And that’s going to be equal to the universal gravitational constant 𝐺 times the mass of Earth divided by the distance between the surface of Earth and the center of Earth, which is the same as Earth’s radius. So let’s call that 𝑟 sub e squared.
Now, the ratio of the local acceleration due to gravity at the height of a geostationary satellite to that on the surface of Earth is equal to 𝑎 sub g divided by 𝑎 sub e. So that’s 𝑎 sub g, which is 𝐺𝑚 over 𝑟 sub g squared, divided by 𝑎 sub e, which is 𝐺𝑚 over 𝑟 sub e squared. Now the simplest way to divide one fraction by another is to invert this term here, which will change this to a multiply. In other words, 𝑎 sub g divided by 𝑎 sub e is equal to 𝑎 sub g times one over 𝑎 sub e. And one over 𝑎 sub e is 𝑟 sub e squared divided by 𝐺𝑚. Now notice here that we have 𝐺 times 𝑚 in both the numerator and the denominator. So those two cancel out, and we’re left with the ratio of 𝑎 sub g to 𝑎 sub e as 𝑟 sub e squared divided by 𝑟 sub g squared.
Now 𝑟 sub e is just a radius of Earth, which we know is 6,370 kilometers. But 𝑟 sub g is the distance between the satellite and the center of Earth. So we need to take the height of the satellite above the equator plus the radius of Earth. So we have 6,370 kilometers squared divided by 35,786 kilometers plus 6,370 kilometers all squared. Now, ordinarily, when we have units of kilometers, we might want to convert that to meters. But here we’re taking a ratio where we have kilometers in both the numerator and the denominator. So the units will cancel out. And in this case we can leave them as they are.
So we can evaluate this to find a value of 0.0228. And we’re asked to give this to three decimal places, so that becomes 0.023. And note that this has no units because it’s the ratio of two quantities with the same units. So the answer to the question “What is the ratio of the local acceleration due to gravity at the height of a geostationary satellite to that on the surface of Earth?” is 0.023.
