help with physics question!?

You don't need the mass of the moon or Earth, just as long as they don't change. Just use the fact that period^2 is proportional to orbit size^3.

Right now (using the numbers in the problem), we have 27^2 is proportional to 1 lunar distance^3.

In 50 billion years, the period will be 1.7407 times longer. Square that, it is 3.030. Cube root is 1.4471. So the distance will be 1.4471 times further than today.

Next, you need to consider that this is the distance to the Earth-Moon center of gravity. Since the moon is 1/80th of the Earth's mass, when the moon moves to 1.4471 d(today), the Earth moves in the other direction by 0.4471 / 80. So the Earth - Moon distance increases by a factor 1.4471 + 0.4471/80 = 1.45266

(47/27)^(2/3) = 1.4471

d(final) = 1.4471 d(today)

P.S. The actual numbers are 27.321582 days and 81.301. Don't know the precise Earth tidal locked orbit period.

The gravitational force between the Moon and Earth is the centripetal force holding the Moon in orbit.

G M_earth M_moon / d^2 = M_moon v^2 / d

Here v is the speed of the Moon's travel around its nearly-circular orbit of radius d.

v = (2 pi d) / T

G M_earth M_moon / d^2 = M_moon (2 pi d / T)^2 / d

The same equations hold true now (with d_initial, v_initial, and T_initial) and in the future (with d_final, v_final, and T_final).

Solve for d_final.

It is possible to express d_final only in terms of d_initial, T_final, and T_initial. This is preferable because those values are much easier to measure than the constant G and the planetary masses.

All of the above assumes that Earth does not move. The equations get just a little more complicated if we take into consideration the gravitational pull of the Moon on the Earth. Now the radius of the Moon's orbit is not the same as the Earth-Moon distance that determines gravitational force, but instead the distance from the moon to the center of mass.

R = 80/81 d

G M_earth M_moon / d^2 = M_moon (2 pi R / T)^2 / R

Re: Additional Details

Each equation is true for any set of measurements taken at the same time.

This means that "v = (2 pi d) / T" actually represents two equations:

v_initial = (2 pi d_initial) / T_initial

v_final = (2 pi d_final) / T_final

You can use keplers third rule.

Treat the system as 1 body containing the mass of both bodies,

with a negligable mass satellite orbiting.

(the result is still valid)

From keplers third rule :

D = cube root ( ( T ² * d ³ ) / t ² )

Key :

D = required distance at 47 day orbit time.

T = 47 days

d = 3.844 e8 meters (existing distance

t = existing orbit time (approx 27.285885 days)

D = 5.52361 E8 meters