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THE “SATELLITE EQUATION” AND KEPLER’S THIRD LAW REVISITED The “Satellite Equation” A satellite may be considered any object which orbits a much larger object by virtue of the gravitational interaction between the two. The Earth has both a very large “natural satellite” (the Moon) and a multitude of much smaller manufactured satellites (communication satellites, the Hubble telescope and the international space station, for example). At the same time, the Earth is itself a satellite of the Sun, as are the other planets in the solar system, along with numerous other objects such as asteroids and comets. Satellite orbits may be either perfect (or nearly perfect) circles or ellipses (highly or only slightly elongated). For simplicity here, however, we will only consider satellites whose orbits are circular. Consider a satellite of mass m in a circular orbit around a planet of mass M (such as the Moon orbiting the Earth). Since the satellite is traveling around a circle at constant speed, it’s centripetal acceleration is v2 a r , 2 and, by Newton’s Second Law, the force exerted on the satellite by the planet can therefore be expressed as mv 2 F ma r . However, because this force is due to the gravitational pull of the planet, it can also be written in another way, according to Newton’s Law of Universal Gravitation, i.e. GMm Fg 2 . r Equating these two expressions for the same force, we can obtain mv 2 GMm 2 , r r and cancelling m and one r from both sides leaves GM v2 r . Finally, taking the square root of both sides, we obtain the “Satellite Equation”, namely: GM v r . Here, it must be remembered that M is the mass of the planet, r is the radius of the orbit and v is the orbital speed of the satellite. 3 The Satellite Equation holds the key to the theoretical behaviour of satellites, and it has a number of extremely important applications. Application 1 Problem: Find the speed of the international space station as it orbits at a distance of 340 km above the Earth’s surface. Solution: First, we find the radius of the orbit to be r 6.38 10 6 340,000 6.72 10 6 m. Then, from the Satellite Equation, v GM 6.67 10 11 5.98 10 24 r 6.72 10 6 m/s. (Note that this is about 27,700 km/h!) Essentially, the satellite equation tells us the exact speed to which we must accelerate a satellite by the time it reaches its orbital radius, and as long as we “shut off the rockets” at that precise radius and speed, a perfect circular orbit will automatically result! In other words, this equation is of critical importance every time a satellite is launched. 4 Application 2 On several occasions already in the course, we have used the fact that the mass of planet Earth is about 5.98 1024 kg. Have you ever wondered how we know the mass of our own planet? In fact, it is thanks to the Satellite Equation and our knowledge of the Moon’s orbit. Problem: Use the Satellite Equation and knowledge of the Moon’s orbit to determine the Earth’s mass. Solution: We know that the accepted radius of the Moon’s orbit is 3.84 108 m. Moreover, we have already used this radius and the time for one complete orbit to calculate the orbital speed of the Moon, which is about 1.02 103 m/s. Squaring and rearranging the Satellite Equation, we have GM v2r v , i.e. M G . 2 r Hence, M 2 v 2 r 1.02 103 3.84 108 11 kg ! G 6.67 10 5 Note that the mass of any astronomical object, which has a satellite whose orbital radius and orbital speed are known, can be calculated in exactly the same way. For example, knowing the Earth-Sun distance and the fact that it takes 365 days for Earth to orbit the Sun is also how we were first able to determine the Sun’s mass. (Since the formula above involves the universal gravitational constant G, however, none of these masses could be determined until after the numerical value of G was found around 1800!) Needless to say, these calculations would be impossible without the Satellite Equation. Kepler’s Third Law Revisited As a final application, recall that Kepler devised and published his three laws of planetary motion well before Newton had invented calculus, and developed Newton’s Laws and the Universal Law of Gravitation. In particular, Kepler’s Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the orbital radius, a result which seemed truly remarkable given that he was able to determine this solely by analysing astronomical data on the movement of the five inner planets. Now, with the help of the Satellite Equation, we can actually prove that Kepler’s Third is correct (at least for circular orbits). 6 The key is to realize that we now have two ways to express the orbital speed of a planet. First, if we call the orbital period T, the speed is just the circumference (i.e. the distance traveled during one orbit) divided by the time, or 2r v T . Second, from the Satellite Equation, GM v r . Equating and inverting these two expressions, T r 2r GM . Squaring both sides, T2 r 4 r 2 2 GM , i.e. 4 2 r 3 4 2 3 T 2 GM r . GM 4 2 Since GM is just a constant for any planet, this proves that, exactly as Kepler proposed, T2 is indeed proportional to r3!

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posted: | 3/21/2010 |

language: | English |

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