Chapter 11 Gravity

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232 Chapter 11 the total energy of each satellite as a function of its mass, the radius of Earth, and its orbital radius. Express the energy difference: Δ E = Egeo − E1000 (1) Express the total energy of an orbiting satellite: Apply Newton’s second law to a satellite to relate the gravitational force to the orbital speed: Solving for v 2 yields: Substitute in equation (2) to obtain: Etot = K + U 2 GM Em 1 = 2 mv − R where R is the orbital radius. F v E radial 2 = tot = GM = R gR R Substituting in equation (1) and simplifying yields: 1 2 2 E E 2 gR m R 2 m v = m R 2 2 2 mgR ⎛ ⎞ E mgRE mgRE ⎜ 1 1 ΔE = − + = − ⎟ 2R ⎜ ⎟ geo 2R1000 2 ⎝ R1000 Rgeo ⎠ 2 mgR ⎛ 1 1 ⎞ E = ⎜ − ⎟ 2 ⎝ RE + 1000 km RE + 35 790 km ⎠ Substitute numerical values and evaluate ΔE: ΔE = 2 E (2) 2 gREm mgR − = − R 2R 6 2 ( 500 kg)( 9.81N / kg)( 6.37 × 10 m) ⎛ 1 1 ⎞ ⎜ − ⎟ = 11. 1GJ The Gravitational Field ( g r ) 2 ⎜ ⎝ 7. 37 × 10 6 m 4. 22× 107 m ⎟ ⎠ 63 •• A point particle of mass m is on the x axis at x = L and an identical point particle is on the y axis at y = L. (a) What is the direction of the gravitational field at the origin? (b) What is the magnitude of this field? 2 E
Gravity Picture the Problem We can use the definition of the gravitational field due to a point mass to find the x and y components of the field at the origin and then add these components to find the resultant field. We can find the magnitude of the field from its components using the Pythagorean theorem. (a) The gravitational field at the origin is the sum of its x and y components: Express the gravitational field due to the point mass at x = L: Express the gravitational field due to the point mass at y = L: Substitute in equation (1) to obtain: r r r g = g + g (1) r g r g x = y = x y Gm iˆ 2 L Gm 2 L (b) The magnitude of g r is given by: 2 2 = g x + g y Substitute for gx and gy and simplify ˆj r r r Gm g g g iˆ Gm = ˆ x + y = + j ⇒ the 2 2 L L direction of the gravitational field is along a line at 45° above the +x axis. ⎛ Gm ⎞ ⎛ Gm ⎞ Gm to obtain: g = ⎜ = 2 2 ⎟ + ⎜ 2 ⎟ 2 ⎝ L ⎠ ⎝ L ⎠ L r 67 ••• A nonuniform thin rod of length L lies on the x axis. One end of the rod is at the origin, and the other end is at x = L. The rod’s mass per unit length λ varies as λ = Cx, where C is a constant. (Thus, an element of the rod has mass dm = λ dx.) (a) What is the total mass of the rod? (b) Find the gravitational field due to the rod on the x axis at x = x0, where x0 > L. Picture the Problem We can find the mass of the rod by integrating dm over its length. The gravitational field at x0 > L can be found by integrating g at x r d 0 over the length of the rod. (a) The total mass of the stick is given by: Substitute for λ and evaluate the integral to obtain: g r M = L ∫ 0 λ dx L 0 2 M = C∫ xdx = 1 2 CL 2 2 233
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Chapter 11 Gravity

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