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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:

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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:

The two components cancels each other

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

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.

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!)

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!)

Use Keppler to find mass of the planet

Then from ag= GM/R^2 find R

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!