The problem is the planet mercury has a radius of 2.44 times 10 to the 6 meters and a mass of 3.30 times 10 to the 23 kilograms. What is the speed of a satellite in orbit? 265 kilometers above mercury surface and what is the period of the satellite? Let's do this. Let's just do given. Here is mercury. Here we have a satellite that is an orbit around mercury. We know the radius that is given and we know the height is given and the problem. We want to find the speed and we want to find the period, capital T. If I go off screen, let me know right away. Let's solve it. You can start off with the orbital velocity equation. I like to start from basic principles. The force of gravity is equal to the universal constant times the mass of mercury times the mass of my spacecraft over the radius of the orbit squared. This force of gravity that Newton told us about is also a centripetal force. Why? Because the satellite is going in a circle around mercury. Centripetal forces are equal to m v squared over r. This mass cancels one of the radius's radii cancel and solve for v. V is the square root of g m over r. It is perfectly okay to start there. We know what g is and what m is that was given. We need to figure out what the radius of our orbit is before we can plug it in there. The radius of the orbit is equal to the radius of mercury plus our height above the surface. The radius was given to be 2.44 times 10 to the 6 meters. The radius of the orbit is equal to the radius of the orbit squared over r. You have to add meters plus meters. You have to take that 265 kilometers and call it 265 thousand meters. You have to calculate the radius of the orbit squared over r. You have to calculate the radius of the orbit squared over r. You have to add that to 265 thousand and you get 2.25. Now we can figure out our orbit velocity 6.67 times 10 to the negative 11, Newton meter squared per kilogram squared times the mass, which is 3.30 times 10 to the 23 kilograms. Over 2.705 times 10 to the 6 meters. The hardest thing about this is entering all those numbers in correctly. Here is my good old HP 15C. The radius of the orbit is equal to the 6.67 times 10 to the 11, negative 11, 3.30 times 10 to the 23. You have to multiply it. Did you get a speed of 2,000? I am going to round it off to 3 significant figures that 2850 meters per second. That is for part A. We know the velocity is equal to 2 pi r over the period. It is a circular orbit. This is the circumference of a circle divided by the time it takes to go around once. That is called the period and that will be equal to our speed. The period is equal to 2 pi r over the speed. That is 2 pi times r, which we have to be 2.705 times 10 to the 6 meters. The period is divided by 2850 meters per second. That gives us an orbital period. Here we have 2 inner pi r over the time it takes to go around twice. We have a time that takes about 5,900 and 60 to the 3 significant figures. We have quite a long time. 60 to the 3 significant figures. I am going to divide that by 3600. It is about 1.65 minutes. Let us take a look at the whole thing. What do you need to know to solve this problem? The radius of the orbit is the height above the planet. The units are the same. You get that answer. Plug in your values. We have an orbital speed of 2850 meters per second. Since it is a circular orbit, we can figure out the period. We have a time that takes about 5,960 seconds. That is it.