# A rock, the earth, an ellipse and the joys of angular momentum

• BigRedRod

#### BigRedRod

I understand that this is a bit cheeky but I've found a problem on one of my mechanics problem sheets that is giving me a headache. As much as I could just ignore it I'd rather try and gain some more understanding of the vague world of rotation and orbit

part 1 said:
A rock of mass m orbits Earth, mass M, radius R with angular momentum L and energy E. Argue directly, considering the direction of motion at its nearest point r to the Earth's centre, that L can written r*sqrt(2m(E + GMm/r))

Even at this early stage I'm bewildered and my searchings and own ponderings still leave me confused. If anyone can quite highlight how L can progress anywhere beyond the obvious r x v (where x is the cross product, and v is the tangental velocity of the rock at it's nearest point to Earth) I'd be forever in their debt

For fun, here is the rest which flows from this first section

part2 said:
Solve for r, and by interpretting the two solutions, deduce the relation e = -GMm/(Ellipse Length)

part3 said:
Suppose the rock is set off by being thrown from the North pole at such a (very fast) speed to have total Enegy E=0 in a direction of our choice.

part4 said:
Conclude (without further calculation) that the escpae velocity is independent of throwing direction

I'm not asking people to do the question for me (although feel free ;) ), any general direction would be pretty damn useful

angular momentum L = mvr

energy E = 1/2(mv^2) - GMm/r

rearrange energy equation until you have v = something...

substitute that into L = mvr for v

V^2 = GM (2/R - 1/a)

The kinetic energy is

K = (1/2) m V^2

K = (GMm/2)(2/R - 1/a)

The gravitational potential energy is

U = -GMm/R

The total energy is

E = K + U

E = (GMm/2)(2/R - 1/a) - GMm/R

E = GMm { (1/2)(2/R - 1/a) - 1/R }

E = -GMm/(2a)

Notice that the expression for the total orbital energy, just above, is the answer to Part 2 of your homework problem.

Solve for the semimajor axis.

a = -GMm/(2E)

The angular momentum L is

L = R x mV

The magnitude of which is

L = RmV sin Q

Where Q is the angle from R to V. Since at periapsis this angle is pi/2 radians, the angular momentum in your homework problem reduces to

L = RmV

Refer again to the Vis Viva equation.

L = Rm sqrt{ GM (2/R - 1/a) }

Substitute the expression for (a) found above.

L = Rm sqrt{ GM (2/R + 2E/GMm) }

Rearrange terms.

L = Rm sqrt{ 2EGM/GMm + 2GM/R }

Simplify.

L = Rm sqrt{ 2E/m + 2GM/R }

Move the orbiting object's mass inside radical.

L = R sqrt{ 2Em + 2GMm^2/R }

Factor out the m.

L = R sqrt{ 2m (E + Gm/R) }

It looks to me as if Part 3 and Part 4 are really one question, and it is a trick question. An orbit with total energy equal to zero is, by definition, barely unbound having exactly the escape speed for its geocentric distance.

Jerry Abbott

Last edited:
Ah, ta muchly guys. That clears everything up