# Central force question?

I know how to use conservation of angular momentum to do part d, but how do I do part e? When will it return to the earth?(What is the criteria?) ### 1 個解答

• 最愛解答

The satellite will return to the planet surface when the graviataional pull on it is GREATER than the required centripetal force to keep it in orbit. That is,

G(m1)(m2)/(r2)^2 > (m2)(Vb)^2/r2 where G is the Universal Gravitational Constant

[Note: when the left term equals to the right term, the satelllite will stay in orbit. If the left term is smaller than the right term, the satellite will fly away from point B]

thus, (Vb)^2 < G(m1)/r2

or (Vb)^2 < G(m1)/X.r1 ------------------ (1)

From conservation of mechanical energy,

loss of kinetic energy of satellite = gain in potential energy of satellite

i.e. (1/2)(m2)(Va)^2 - (1/2)(m2)(Vb)^2 = G(m1)(m2)(1/r1 - 1/r2)

Because r2 = X(r1), we have, after simplification,

(Va)^2 - (Vb)^2 = 2G(m1)(X - 1)/(X.r1)

i.e. G(m1)/(X.r1) = [(Va)^2 - (Vb)^2]/(2.(X - 1))

Substitute into (1):

(Vb)^2 < [(Va)^2 - (Vb)^2]/(2.(X - 1))

Divide both sides by (Vb)^2,

1 < [(Va/Vb)^2 -1]/(2.(X -1))

Inverting the fraction on the right-hand-side, then

2(X -1)/[(Va/Vb)^2 - 1] < 1 --------------- (2)

I suppose you have already found the expression for Vb/Va in part (d). Let this be f, say.

i.e. Vb/Va = f where f is a function of "alpha" and X

Hence, inequality (2) becomes,

2(X -1)/[(1/f)^2 - 1] < 1

The expression on the left-hand-side involves X and "alpha" only. I just leave it for you to complete the derivation, as it involves only algebric manipulation.

Hope the above would help.