Chapter 13 Universal Gravitation Examples. Example 13.1 “Weighing” Earth “Weighing” Earth! : Determining mass of Earth. M E = mass of Earth (unknown)

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Chapter 13 Universal Gravitation Examples

Example 13.1 “Weighing” Earth “Weighing” Earth! : Determining mass of Earth. M E = mass of Earth (unknown) R E = 6,370 km = 6.37 x 10 6 m Assume the Earth is a uniform, perfect sphere. m = mass of object on the surface of Earth (known) The weight of m is: F g = mg The gravitational force on m is: F g = G(M E m)/R E 2 But these two are the same!!! mg = G (mM E )/R E 2  g = G M E /R E 2  Solving for M E M E = g R E 2 /G = 5.98  kg MEME RERE m

Example 13.2 The Density of the Earth (Example 13.3 Text Book) Using the last result, find the average density of the Earth: From Example 13.1 mg = G (mM E )/R E 2 &g = G M E /R E 2  M E = g R E 2 /G Since:  E = M E /V E & V E = (4/3)  R E 3 

Find F g acting on a 2000 kg spacecraft at 2R E R E = 6,380 km & M E = 5.98  kg F G = G(mM E /r 2 ) At surface (r = R E ) F G = mg = G(mM E /R E 2 ) At r = 2R E F G = G(mM E )/(2R E ) 2  F G = G(mM E )/4R E 2  F G = ¼ G(mM E /R E 2 )  F G = ¼mg = 4900 N Example 13.3 Spacecraft at 2R E

Example 13.4 Geosynchronous Satellite A Geosynchronous Satellite appears to remain over the same point on the Earth The gravitational force supplies a centripetal force A satellite of mass m is moving in a circular orbit around the Earth at a constant speed v and at an altitude h above the Earth’s surface. Find the speed (v) of the satellite in terms of G, h, R E and M E

Example 13.4 Geosynchronous Satellite, final Assume m is moving around the Earth in a circular orbit. Applying Newton’s 2 nd Law: Taking (1) = (2): Since: r = R E +h

Example 13.5 The Change in Potential Energy (Example 13.6 Text Book) A particle of mass m is displaced through a small vertical displacement  y near the Earth’s surface. Show that the general expression for the change in gravitational potential energy (Equation 13.12) reduces to the familiar relationship:  U = mg  y From Equation Since r i and r f are to close to the Earth’s surface, then: r f – r i ≈  y and r f r i ≈ R E 2 Remember r is measured from center of the Earth  U = mg  y

Examples to Read!!! Example 13.1 (page 376) Material from the book to Study!!! Objective Questions: Conceptual Questions: Problems: Material for the Final Exam