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51010 Assignment
1)
Length of major axis = 9.2 cm
Length of minor axis = 2 cm.
2) Closest point is the left vertex (-a,0) = (-4.6,0) is the periluna
(4.6,0) is the apuluna
3) Distance of perluna to moon centre = 4.6 cm
Because focus is symmetrical about minor axis we get the other focus to the left of apuluna by a distance of (a-ae) units where e = eccentricity
Eccentricity = =0.9760
Focus would be at a distance of 0.9760*4.6 from origin on either side
i.e. (4.4896,0) and (-4.4896,0) are the two foci.
4) Orbit eccentricity = focus/a=4.4896/4.6 = 0.9760
To check whether this orbit obeys Kepler’s 3rd law:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Since orbital path is ellipse we have as per ellipse rule
Eccentricity is constant always.
i.e. Period = 1/distance = 1/(a-ae)
a-ae = a-
(P-a)^2 = a^2(a^2-b^2)
i.e. , which is true by definition of ellipse
Thus we get P^2 = a^3 is satisfied
Kepler second law states that the satellite’s motion should sweep out equal areas in equal times.
1)
Let us consider these triangles areas one by one
The parametric form of ellipse would be
X = a cost and y =a sin t
So base of this triangle between (acost1, asint1 ) and (acost2, bsint2) would be
Base ^2 = (acost1-acost2)^2 +(bsint1-bsint2)^2
Height = distance of origin from line connecting the two points
The line connecting two points is
y-bsint2 =
(acost1-acost2) (y-bsint2) –(bsint1-bsint2)(x-acost2) =0
Distance can be calculated using distance formula for a point from a line
D^2 =
Similarly for other two points t3, and t4 we arrive them
Since t1-t2 = t3-t4 by simple calculation we can find
The value b^2h^2 remains constant.
Hence ½ bh = area of any triangle would be the same for equal periods.
3) The Moon is actually going at its maximum speed when it it is at minimum distance from Earth but it the slows down until it reaches it apogee.
Hence at Periluna...