Question

(b) [6] For a more realistic case in which the comet experiences gravitational forces, consider a comet approaching Earth. Estimate the distance from the Earth (in Earth radii) at which Earth’s gravity begins to be as important as, or more important than, that of the Sun. Neglect the Moon’s gravity. Discuss the effect of the Moon’s gravity.

ps: i already have answer for part a but for more background on the question, part a is listed below

a.Suppose that a comet,5km in diameter,is moving directly towards Earth at a constant speed of 10km/s. On one night (call this t = 0), its angular diameter subtends an angle of 2 arcseconds. Assuming that its speed remains constant, how long will it take before its angular diameter subtends an angle of 2 arcminutes? Note that by assuming the comet moves at a constant speed, this problem explicitly neglects any gravitational forces.

Answer 1

b. let the distance of the comet from earth be x
now, distance of earth from sun , d = 151.93*10^9 m
radius of earth, r = 6.371*10^6 m
mass of earth , m = 5.972*10^24 kg
mass of sun, M = 1.989*10^30 kg

hence, x at which force of sun = force of earth due to gravity on comet
assuming that the comet , earth and the sun are collinear with distance between sun and comet = x + d
hence

GM/(x + d)^2 = Gm/x^2
x^2/m = (x + d)^2/M
sqrt(M)*x = sqrt(m)(x + d)
x = sqrt(m)*d/(sqrt(M) - sqrt(m))
x = 0.2637176722701897511512985278212*10^9 m
in earth radii, x = 41.39 R [ where R is earth radius]
so for distance less than x = 41.39R, the gravity of earth is greater on the comet than sun

at this distance, gravity due to moon is 0.07346/5.9724 times the gravity of earth on comet ( because mass of moon is 0.07346/5.9724 times mass of earth)
hence, gravity due to moon is 1.2299% of gravity of earth, hence its not important at this distance

a. given, d = 5 km
v = 10 km/s
at t = 0
theta = 2 arc seconds = 2/60 arc minutes = 2/3600 deg = 1/1800 deg = pi/180*1800 rad
now theta = d/x = 5/x
x = 5*180*1800/pi = 515662.0156177 km is distance of comet from earth at t= 0

now at t = 0, theta = 2 arc min = 1/30 deg = pi/180*30 rad
hence
x2 = 5*180*30/pi = 8594.3669269623 km at t= t
hence distance travelled by comet = x1 - x2 = 507067.648690 km

hence time taken for the comet to cover this distance is 507067.6486 km / 10 km/s = 50706.764869 s
t = 845.112 min = 14.085 hours

hence the comet will subtend an angle of 2 arc min by the next day afternoon