In this problem we study two small satellites (masses m, and m,) orbiting the Ea

Question

In this problem we study two small satellites (masses m, and m,) orbiting the Earth. The mass of the Earth, M, is so large that we can assume that the center of the Earth remains at the origin of our xyz coordinate system. The North Pole is on the positive z axis. The coordinate system does not rotate. The satellites have negligible gravitational interaction with each other, but they do collide and stick together at time to (a) Before the collision, the satellite with mass m, is in a polar orbit, meaning it passes over the North and South poles. The orbit is circular with radius R. Calculate the speed va of the satellite in terms of G, M and R. (b) Before the collision, the satellite with mass m2 is in an equatorial orbit, meaning the orbit is in the xy plane. The orbit is elliptical with rmin R and mx 2R. Calculate Vb, the speed of the satellite at rmin. Calculate the eccentricity of the orbit. Calculate the period using a theorem; state the theorem and the main ideas in its proof. (c) The location of the collision is (R, 0, 0). Just before the collision, the velocity vector of mi points in the +z direction, and the velocity vector of m2 points in the ty direction. What is the velocity vector of the combined object immediately after the collision? In your answer use the symbols Va and vb. You do not have to plug in the results you got earlier for these speeds. (d) The remaining questions are about the orbit of the combined object after the collision. Compute a vector perpendicular to the plane of the orbit. (e) Compute the eccentricity of the orbit.

Explanation / Answer

For the given problem

a. before the collision, the satellite with mass m1 , is in polar orbit

circular

radius R

hence

let its speed be va

then from force balance

GM*m1/R^2 = m1*va^2/R

GM/R = va^2

va = sqroot(GM/R)

b. the orbit of m2 is in xy plane

elliptical orbit

rmin = R

rmax = 2R

vb = velocity at rmin

hecne

vb = sqroot(GM(2/rmin - 1/rmax))

vb = sqroot(GM(2/R - 1/2R)) = sqroot(GM*3/2R)

vb = sqroot(3GM/2R) = sqroot(3/2)va

c. location of collision (R,0,0)

just before collision, va is in +z direction

vb is in +y direction

immidiately after collision

v = va + vb

v = va*k + sqroot(3/2)va*j

v = sqroot(GM/R)(k + sqroot(3/2)j)

hence velocity just after collision |v| = sqroot(5GM/2R)

d. a vector perpendicular the the plane of the orbit = r

r' = vector parallel to v

r' = (sqroot(3/5)j + sqroot(2/5)k)

r' = cos(39.23)j + sin(39.2315)k

hence

r'= cos(39.2315 + 90) j + sin(39.2315204 + 90)k

r' = -0.6324555j + 0.7745966 k

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