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CH13.Problems

CH13.Problems. JH. 132. theta: to x-axis: tan theta = 3/4  theta = 36.87 R = Sqrt (3^2+4^2) = 5m G = 6.67*10^-11 N m^2/kg^2 Forces along x: F= - 2*(G m M/R^2 cos (theta) ) = -2*1000*10000* 6.67*10^- 11* cos (36.87)/5^2 = -4.3*10-5 N Forces along y axis:

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CH13.Problems

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  1. CH13.Problems JH.132

  2. theta: to x-axis: tan theta = 3/4  theta = 36.87 R = Sqrt(3^2+4^2) = 5m G = 6.67*10^-11 N m^2/kg^2 Forces along x: F= - 2*(G m M/R^2 cos(theta) ) = -2*1000*10000* 6.67*10^-11*cos(36.87)/5^2 = -4.3*10-5 N Forces along y axis: The two components cancels each other

  3. Pay attention to altitude or above surface. You must add R to find distance from center of Earth

  4. The computed speed is escape velocity of Earth

  5. Mass of the satellite has no effect on period.

  6. Notice how Kepler law is used to compute the mass of Mars. Just observe satellites for their periods and orbit radii.

  7. The final potential energy is identical to initial potential energy.

  8. Need to find force of B& C on A, then Make force of D cancel force of B&C: Find AB and AC force; then find its magnitude and direction reverse direction then: Find components of x and y for the new force assuming position of (xd, yd) and 4ma Solve for x and y component wise.

  9. Make the force of gravity equal to uniform rotation acceleration: GMm/r^2 = m v^2/r - GM/r^2 = w^2 * r M= w^2* r^3/G = 5x10^24 kg (Earth mass in 20 km!)

  10. Use Newton’s Spherical shell theory: • Outside the shell: shell is a point at the center • Inside the shell: shell has zero effect • GMm/r^2 = m v^2/r - GM/r^2 = w^2 * r • M= w^2* r^3/G = 5x10^24 kg (Earth mass in 20 km!)

  11. Use Keppler to find mass of the planet Then from ag= GM/R^2 find R

  12. Find energy cost to change the potential energy Uf-Ui Find kinetic energy at orbit: ½ m Vorbit^2 Find h at which the above are equal b) When h of greater than h above: WHY USA and Russia put their space Launching as close as possible to the Equator!

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