Classical Dynamics of Particles and Systems
5th Edition
ISBN: 9780534408961
Author: Stephen T. Thornton, Jerry B. Marion
Publisher: Cengage Learning
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Suppose one of the Global Positioning System satellites has a speed of 4.34 km/skm/s at perigee and a speed of 3.20 km/skm/s at apogee.
If the distance from the center of the Earth to the satellite at perigee is 2.43×104 kmkm , what is the corresponding distance at apogee?
If you wanted to place a satellite in a geosynchronous orbit, at what height above the Earth's surface should you place it? (Earth's mass = 5.98x10^24, Earth's radius = 6x10^6, G = 6.67x10^-11)
For the previous situation using the given answer, what is the difference in potential energy between the Earth's surface and the satellite height for a satellite with 1 kg mass?
One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R, and mass 4M. If M= 4.1 * 1024 kg and R = 6.0 *106 m, what is the gravitational acceleration of a particle at points (a) R and (b) 3R from the center of the planet?
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- An Earth satellite has its apogee at 2500 km above the surface of Earth and perigee at 500 km above the surface of Earth. At apogee its speed is 730 m/s. What is its speed at perigee? Earth’s radius is 6370 km (see below).arrow_forwardUsing Figure 13.9, carefull sketch a free body diagram for the case of a simple pendulum hanging at latitude lambda, labeling all forces acting on the point mass,m. Set up the equations of motion for equilibrium, setting one coordinate in the direction of the centripetal accleration (toward P in the diagram), the other perpendicular to that. Show that the deflection angle , defined as the angle between the pendulum string and the radial direction toward the center of Earth, is given by the expression below. What is the deflection angle at latitude 45 degrees? Assume that Earth is a perfect sphere. tan(+)=gg2REtan , where is the angular velocity of Earth.arrow_forwardA Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 40,000.0 km as measured from the center of Earth and a velocity of 3.0 km/s, what would be its velocity at perigee measured at 200.0 km altitude?arrow_forward
- Given the perihelion distance, p , and aphelion distance, q , for an elliptical orbit, show that the velocity at perihelion, vp , is given by vp=2GMSun(q+p)qp . (Hint: Use conservation of angular momentum to relate vp and vq , and then substitute into the conservation fo energy equation.)arrow_forwardShow that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartersian coordinates, x and y, for the polar coordinates, r and , and showing that it has the general form for a parabola, x=ay2+by+c .arrow_forwardA satellite in a geosynchronous circular orbit is 42,164.0 km from the center of Earth. A small asteroid collides with the satellite sending it into an elliptical orbit of apogee 45,000.0 km. What is the speed of the satellite at apogee? Assume its angular momentum is conserved.arrow_forward
- Model the Moons orbit around the Earth as an ellipse with the Earth at one focus. The Moons farthest distance (apogee) from the center of the Earth is rA = 4.05 108 m, and its closest distance (perigee) is rP = 3.63 108 m. a. Calculate the semimajor axis of the Moons orbit. b. How far is the Earth from the center of the Moons elliptical orbit? c. Use a scale such as 1 cm 108 m to sketch the EarthMoon system at apogee and at perigee and the Moons orbit. (The semiminor axis of the Moons orbit is roughly b = 3.84 108 m.)arrow_forwardCalculate the effective gravitational field vector g at Earths surface at the poles and the equator. Take account of the difference in the equatorial (6378 km) and polar (6357 km) radius as well as the centrifugal force. How well does the result agree with the difference calculated with the result g = 9.780356[1 + 0.0052885 sin 2 0.0000059 sin2(2)]m/s2 where is the latitude?arrow_forwardCircular orbits in Equation 13.10 for conic sections must have eccentricity zero. From this, and using Newton’s second law applied to centripeta acceleration, show that the value of in Equation 13.10 is given by Where is the angular momentum of the orbiting body. The value of is constant and given by this expression regardless of the type of orbit.arrow_forward
- It was stated that a satellite with negative total energy is in a bound orbit, whereas one with zero or positive total energy is in an unbounded orbit. Why zero or positive total energy is in an unbounded orbit. Why is this true? What choice for gravitational potential energy was made such that this is true?arrow_forwardIf a spacecraft is headed for the outer solar system, it may require several gravitational slingshots with planets in the inner solar system. If a spacecraft undergoes a head-on slingshot with Venus as in Example 11.6, find the spacecrafts change in speed vS. Hint: Venuss orbital period is 1.94 107 s, and its average distance from the Sun is 1.08 1011 m.arrow_forwardA geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are sueful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for Earth in Appendis D.arrow_forward
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