According to Kepler\'s first law, a comet should have an elliptic, parabolic, or

Question

According to Kepler's first law, a comet should have an elliptic, parabolic, or hyperbolic orbit (with gravitational attractions from the planets ignored). In suitable polar coordinates, the position (r. 8) of a comet satisfies an equation of the form r= + e(r-cos ), where is a constant and e is the eccentricity of the orbit, with 0 se1 for an ellipse, e 1 for a parabola, and e>1 for a hyperbola. Suppose observations of a newly discovered comet provide the data below. Determine the type of orbit, and predict where the comet will be when = 4.2 (radians) 90.83 1.091.421.762.17 r 3.52 3.17 1.99 1.06 0.63 The comet has a hyperbolic orbit. When 8 = 4.2 (radians), the comet will be at r (Round to two decimal places as needed.)

Explanation / Answer

r=B+e(r*cos(theta))...(i)

Values of r and theta are given in the table so let's plug last two points (1.76,1.06) and (2.17,0.63) in equation (i) to find values of B and e

using (1.76,1.06), we get:
1.06=B+e(1.06*cos(1.76))
or 1.06=B+e(1.06*-0.188)
or 1.06=B-0.19928e
or 1.06+0.19928e=B...(ii)

using (2.17,0.63), we get:
0.63=B+e(0.63*cos(2.17))
or 0.63=B+e(0.63*-0.564)
or 0.63=B-0.35532e
plug value of B from (ii)
0.63=1.06+0.19928e-0.35532e
0.63-1.06=-0.15604e
-0.43=-0.15604e
-0.43/-0.15604=e
2.75570366573=e

plug value of e into (ii)
B=1.06+0.19928e=1.06+0.19928*2.75570366573=1.60915662651

Now plug values of B and e into (i) we get equation:
r=1.60915662651+2.75570366573(r*cos(theta))

Now we have to find value of r when theta=4.2
r=1.60915662651+2.75570366573(r*cos(4.2))
r=1.60915662651+2.75570366573(r*-0.49)
r=1.60915662651-1.35029479621r
2.35029479621r=1.60915662651
r=1.60915662651/2.35029479621
r=0.684661613132
which is approx 0.68
So the final answer is r=0.68

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