The references:

Newton's Law of Gravitation shows that to find the force of gravity (Fy) between Earth and any other object, we can use =% where m, represents the mass of Earth, m represents the mass attracted by Earth, and R, represents the distance between the object and the g th centre of Earth, in other words, the radius of Earth if the object is on the surface of Earth. We also know from other work that the force of gravity between Earth and the mass is given by Fy; = mg where g is the acceleration due to gravity at the surface of Earth, or the gravitational field constant for Earth. Therefore Gm,m =mg R; Solving for "g", g= Gm, Rcz We can thus determine the value of g at various heights above the earth. For example, we can determine the effective value of g at a height of 1000 km above the surface of Earth. In doing this, we must be very careful about the value we use for the radius. To determine this we add together the actual radius of Earth to the height above Earth. R =6380 km + 1000 km = 7380 km = 7.38 x 10 m (6.67x107"'N - m*/kg %)(5.98x10%*kg) = = =7.32 N/kg or 7.32 m/s? (7.38x10m)> This method of determining g could be used to calculate its value for any planet once the mass of the planet and the radius of the planet are known. The value of g could then be used to determine the weight of an object using Fy = mg. The equation g= Gn:zc is also useful for measuring the mass of a planet. When Cavendish determined the value of G, he was also the first to calculate the mass of the earth. The value of g was known, as was the radius of the earth. It then became possible to determine the mass of the earth using 2 2 6 2 y 8RE_ OB0M/S)638XI0M' _ oo oy G 6.67x10""'N- m'kg' Similarly the mass of any planet can be calculated if the value of the gravitational field constant, g, and the radius, R, of the planet are known. as Newton's Thought Experiment and Satellite Motion Long before man was able to place artificial satellites into orbit, Newton thought about the conditions for satellite motion. Suppose you were on top of a mountain, and you threw a stone horizontally. The stone will move in a parabolic path characteristic of projectile motion and strike the earth (path A below). If the stone is thrown with a greater horizontal velocity, it will travel further before striking the earth (path B below). If the stone were thrown with just the right speed, it would not strike the ground at all but move around the circular in a circular orbit (path C below). A B /' Newton was also able to calculate how fast an object would have to be launched horizontally for it to move in a circular orbit. The answer depends on the rate at which the stone falls and the rate at which Earth curves. When any object is dropped from rest, it accelerates downwards at 9.80 m/s? and falls a distance of 4.9 m in 1.0 s. The same is true of a stone thrown horizontally. A geometric fact about Earth is that its surface drops a vertical distance of nearly 5 m for every 8000 m tangent to the surface. This means that if your eyes were level with a calm water surface, you would be able to see only the very top of a 5 m tall mast on a ship that is 8 km away. Thus, if a stone could be thrown so that it travels 8 km in 1 s, then it will fall 5 m, and it will follow the curvature of Earth. It's horizontal speed is 8 km/s which is equivalent to 29 000 km/h. At this speed, an object will burn up in the atmosphere. Artificial satellites circling the earth are now commonplace. A satellite is put into orbit by accelerating it to a sufficiently high tangential speed with the use of rockets. If the speed of the satellite is too high, the spacecraft will not be confined to Earth's gravity and will escape, never to return. If the speed is too low, it will fall back to Earth. Satellites are usually put into circular, or nearly circular orbits. What keeps the satellite up is its high speed and the gravitational force of Earth keeping it in its orbit. For satellites that move with uniform circular motion, the centripetal acceleration is given by _. The force that gives a satellite this acceleration is the force of R gravity. The gravitational force supplies the centripetal force necessary to keep the satellite in orbit. Thus we can write mme v 2 R2 R In this equation, R refers to the distance of the satellite from Earth's centre, and v is the speed of the satellite in its orbit. We can thus calculate the speed that the satellite must have using Gme V = R For example, if a satellite orbits Earth at a height of 200.0 km above Earth's surface, then the velocity it must have to stay in orbit is (6.67x10-'N . m'/kg2)(5.98x102*kg) V = 1 =7.79 x 103 m/s 6.38x10'm+ 2.000x10'm Because the satellite is moving in a circular orbit, we know that the time it takes the satellite to travel once in its orbit is T = 2ntR 2n(6.58x10'm) = 5.31 x 103 s = 88.5 min V 7.79x10'm/s It is also possible to determine the radius of orbit of a satellite if the period of motion of the satellite and the mass of the central body are known. Using Earth as the central body, 2TR m me v2 T 412 R R2 = ms R =m s = ms R T2 Thus R=3/ Gm T2 4TC2-+ Mass of a Central Object and Radius of Motion of a Satellite It is also possible to determine the mass of a central object such as a planet or a star if the speed of any satellite orbiting this central is known. If mg represents the mass of the orbiting satellite, and m represents the mass of the central object, then Notice that the mass of the satellite is irrelevant to determining the mass of the central object. To determine the mass of the central object using the radius of motion of the satellite's orbit, and the period of motion of the satellite, remember that for circular motion, 2nR v= T Substituting this into the equation for mass, 2 2R o T ) 4R G GT? One of the interesting applications of this result is that we can determine the mass of any planet in our solar system if we can measure the radius of orbital motion and the period of motion of any satellite, such as a moon, circling the planet. We can easily determine the mass of our own sun by knowing the mean radius of orbit and the period of motion of our own Earth. IMAGE we SUMMary pixels The value of the gravitational field constant for any planet of mass m and radius R can be determined using _Gm = R2 This equation can also be used to determine the value of g at any distance R from the centre of the planet. If the value of g and the radius R of a planet are known, then the mass of a planet can be determined using gR' m== G Newton reasoned that if an object were launched horizontally at a fast enough speed, that it would orbit Earth. He was able to calculate this speed by knowing the radius of Earth and the rate at which objects accelerate when they fall. By using our knowledge of circular motion and Newton's law of universal gravitation, the velocity a satellite must have to stay in orbit is v = Gm V R 2nR its period of motion is 77 = Z~ v and the radius of orbit is R=3 GmT? \\l 4m? where m is the mass of the planet, R is the radius of orbit, T is the period of revolution, and v is the speed of the satellite. If the radius or orbit and the period of orbit of a satellite are known, then the mass of the central body around which the satellite moves can be determined using _ 4R' GT* m Question(s): The physics of satellite motion around Jupiter. A satellite of mass 2.00 x 10% kg is placed in orbit 6.00 x 10 m above the surface of Jupiter. Please refer to the data table for planetary motion included in this lesson. 1. Determine the force of gravitational attraction between the satellite and Jupiter. 2. What must be the orbital speed of the satellite? 3. What must be the value of the gravitational field constant, g, at the location of the satellite? 4. One of the moons of Jupiter is Europa. It's period of motion is 3.07 x 10s. What must be the radius of its orbit? 5. If a satellite was placed in orbit around Europa at a height of 100.0 km above the surface of Europa, and period of motion was 7.58 x 103 s, what must be the mass of Europa? The radius of the moon Europa is 1.57 x 106 m