Circular and Satellite Motion Legacy Problem #22 Guided Solution

The International Space Station is an orbiting satellite that orbits the Earth at a height of about 350 kilometers above Earth�s surface. We�re asked to determine the speed of the International Space Station. We often think of the speed of an orbiting object being the 2 pi r divided by the period. Here we�re given the radius, or at least sufficient information to determine the radius of the orbit, but we�re not given the period. We are not, for in the overview page of this set of problems, you�ll notice there�s another equation relating the orbital speed of a satellite to the radius of orbit and the mass of the object about which it orbits, here being the Earth. That equation states that the speed of an orbiting satellite is the square root of big G times the mass of the object being orbited divided by the radius of the orbit. So if we can calculate the Gm per r and then take the square root of it, we have the speed of our orbiting space station. Now one complication in this problem is that the r, the radius of orbit, is the radius of orbit. It�s not the radius of Earth and it�s not the 350 kilometers altitude, but it�s the actual radius of orbit, which can be found by taking the radius of the Earth and adding to it the altitude or height of the space station above Earth�s surface. If you were to draw a picture of the Earth and then a picture of the satellite or the space station, you�d put it a small distance above Earth�s surface. And then if you think of it orbiting Earth, it maintains this small distance of altitude above Earth�s surface. And so the actual radius of its orbit would be a little bit greater than the radius of the Earth. To determine this radius of orbit, we have to take the 350 kilometers and first convert it to meters, since we�re given the radius of the Earth in meters. So we multiply by 1000 to get 3.5 times 10 to the 5th meters and then add that on to the radius of the Earth. This gives us a figure of 6.72 times 10 to the 6th meters. Now that�s the denominator inside the square root symbol. In the numerator we go big G, 6.673 times 10 to the negative 11th, multiplied by the mass of the Earth here given as 5.98 times 10 to the 24th. We divide by this radius of orbit and then we take the square root of the whole thing. We get about 7686.6 meters per second. We can round that to three significant digits and so we would get 7.69 times 10 to the 3rd meters per second.