Circular and Satellite Motion Legacy Problem #27 Guided Solution

A good problem solver reads the problem carefully and identifies the known quantities and the unknown quantities, and then thinks about physics concepts and math relationships in order to plot out a strategy to get from the known quantities to the unknown quantities. Here we're given a story of a spacecraft that's orbiting the planet Mercury. We're told that it orbits Mercury at a height of 125 miles above the surface of Mercury. We're given the radius of Mercury, and the mass of Mercury, and some conversion information in order to convert the miles to units of meters. Now as I read this problem, I immediately began to draw a picture of a planet. I drew a circle, I labeled it Mercury, and then I located the satellite, which was not on its surface obviously, but above its surface, a little bit above its surface, and it was orbiting in what I drew as a circular path. Now what I'm asked to do here is to calculate the speed in the orbital period. Now I know that the speed of orbit depends upon the mass of the planet being orbited, and it depends upon the radius of orbit. So I can now use my diagram to try to figure out the radius of orbit. I know that the distance from the center of this inner circle, the planet circle, to the edge of the surface of Mercury, I know that that distance is 2.44 times 10 to the 6 meters, we call that the radius of the planet. But I know that the radius of the orbit is a little bit bigger than that, by 125 miles, since the distance from the surface of Mercury out to where the planet, where the satellite is at, is 125 miles. So what I have to do is add 125 miles to 2.44 times 10 to the 6 meters in order to get the radius of orbit. Now miles don't add very easily to meters, and so my first step is to take 125 miles and multiply by 1609 meters per mile in order to get the height of the satellite above the surface of the planet in units of meters. I end up getting 2.01125 times 10 to the 5th meters, and I can add that onto the radius of Mercury to get the radius of orbit, and that gives me about 2.641125 times 10 to the 6 meters, and not all those digits are significant, but I'll round it when I get to the final answer. Now, I'm going to record that number, because I have to use it a couple times here, I have to use it to calculate orbital speed, I have to use it to calculate orbital period, and if I go to the set, the orbit view page for the set of problems, one thing that I'll notice is that there's an equation that relates this orbital speed to the mass of the planet being orbited and the radius of orbit. Goes something like this, V equal big G times mass of Mercury divided by the orbital radius and the square root of the whole thing. So I'm just going to take 6.673 times 10 to the negative 11 for big G, the mass of Mercury, I'm going to multiply those two things together, I'm going to divide by this radius of orbit I just calculated, I'm going to find out the result, and take the square root of the whole thing. I get about 2887.5 meters per second, I'm going to round that to three significant digits and it becomes 2.89 times 10 to the 3rd meters per second, but I'm going to take the unrounded number for speed and use it in my next calculation for the orbital period, because the orbital period is related to orbital speed and radius of orbit by the equation V equal 2 pi r divided by period. I can rearrange that to get period by itself, so period is 2 pi r divided by the speed. The r I've calculated to be 2.641125 times 10 to the 6th meters and hopefully we're down, and the speed I just calculated to be 2887.5 meters per second, and hopefully you still have that on your calculator, so I'm going to go 2 times pi times radius divided by the speed, and that will give me about 5747 seconds, I can round that to three digits, 5.75 times 10 to the 3rd seconds, that's the orbital period. If you think about that in terms of us observing it from Earth, we would observe one orbital period of the satellite about Mercury every about 1.6 hours or so.