<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
Hello!
Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:
← The circumference of the orbit
speed = orbital speed, we will solve for this later
time = period
Therefore:
Where 'r' is the orbital radius of the satellite.
First, let's solve for 'v' assuming a uniform orbit using the equation:
G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)
m = mass of the earth (5.98 × 10²⁴ kg)
r = radius of orbit (1.276 × 10⁷ m)
Plug in the givens:
Now, we can solve for the period:
