For a satellite to be in a circular orbit 890 km above the surface of the earth, (a) what orbital speed must it be given?

Video transcript

Welcome back everybody. So four global positioning systems or for GPS satellites, we have satellites that orbit at an average height of 20,200 kilometers above the Earth's surface. Now this in meters is just two point oh two times 10 to the seventh meters. It's easier to work with there. And we are asked to find what the orbital speed is of the satellites to maintain this orbit. But we know from kepler's law that the orbital speed is equal to the square root of Newton's gravitational constant times the mass of whatever body is being orbited, in this case Earth all over the radius of orbit. But what is the radius of orbit? Is that our heights? Well not quite radius of orbit is going to be the distance of the satellite away from the center of earth, meaning that the radius of the orbit is equal to the radius of Earth plus the height of the orbit from the Earth's surface. That makes sense by our little diagram over here. So let's go ahead and find our real quick, right? So r is equal to the radius of Earth, which you can just look it up. It's it's to be 6.37 times 10 to the six m plus our height of orbit, which is 2.2, two times 10 to the seventh meters. This gives us a radius of orbit is 2.657 times 10 to the seventh meters. Great. Now that we have that we can go ahead and plug it into this equation right here to find our orbital speed. we have that velocity is equal to square root of 6.67 times 10 to the negative 11 times the mass of Earth, which you can also look up, which it is 5.97 times 10 to the 24th kilograms, all divided by our radius of orbit of 2.657 times 10 to the seventh, giving us a final answer of 3870 m per second corresponding to answer choice. C Thank you all so much for watching. Hope this video helped. We will see you all in the next one.