ch13.problems
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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: - PowerPoint PPT PresentationTRANSCRIPT
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CH13.Problems
JH.132
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theta: to x-axis: tan theta = 3/4 theta = 36.87R = Sqrt(3^2+4^2) = 5mG = 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
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Pay attention to altitude or above surface. You must add R to find distance from center of Earth
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The computed speed is escape velocity of Earth
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Mass of the satellite has no effect on period.
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Notice how Kepler law is used to compute the mass of Mars. Just observe satellites for their periods and orbit radii.
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The final potential energy is identical to initial potential energy.
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Need to find force of B& C on A, thenMake 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 4maSolve for x and y component wise.
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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!)
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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!)
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Use Keppler to find mass of the planet
Then from ag= GM/R^2 find R
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Find energy cost to change the potential energyUf-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!