Question

Venus rotates slowly about its axis, the period being 243 days. The mass of Venus is 4.87x10^24 kg. Determine the radius for a synchronous satellite in orbit about Venus.

Answer 1

@baldymcgee6

Answer 2

eq1 Fg = Gm1m2/r^2
eq2 Fa = m2 v^2/r

G is the gravitational constant

m1 is the mass of venus (4.87x10^24 kg)

m2 is that mass of your satellite. Its really a dummy variable that you can throw out. (I'll explain why soon)

Fg is the force of gravity between two objects, and Fa is the "centripetal" acceleration force felt by the sattelite. Since these forces have to be the same for something to stay in orbit.

You know that force is mass*acceleration. Since Fg=F=Fa, if you divide both sides by m2, you will get accelerations, which also happen to be equal since the forces are divided by the same mass i.e., ag = A = aa. (F = m2 * A ...)

So you have
(A = Gm1/r^2 = v^2/r) * r^2
= Gm1 = v^2 * r

You are stuck with 1 eq but 2 unknowns, right? Nope!
v = r w, where w is the rotational speed (2pi/243days)

so, you have

= Gm1 = w^2 * r^2 * r
= Gm1/w^2 = r^3

Therefore:

r = (G*m1/(w^2))^(1/3)

and make sure that when you plug in, that you do everything in units of meter-Kg-second.

GM/r^2 = (1/2)v^2/r = (1/2)r^2ω^2/r = (1/2)r(4π^2)/P^2 = r(2π^2)/P^2
r^3 = GMP^2 / (2π^2)
r = [(6.67428*10^-11 m^3/kg-s^2)(4.87*10^24 kg)((243 da)(24 hr/da)(3600 s/hr))^2 / (2π^2)]^(1/3)
r = 1,936,190 km ≈ 1,203,092 mi.

Answer 3

@ksaimouli

Answer 4

time period required=243 days=20995200sec

taking h to be negligible or ver small as compared to R we get T:)