If the sun is 372 times farther from the Earth than the Moon, and the Moon is falling around the earth at 1.4 mm/s, what is the rate that the Sun is "falling?"

Question

If the sun is 372 times farther from the Earth than the Moon, and the Moon is falling around the earth at 1.4 mm/s, what is the rate that the Sun is "falling?"

Is it 1/(372^2) or 7.22*10^-6 mm/s?

Answer 1

I guess by falling it means tangential velocity

F = ma = mv^2/r where m is mass of body n orbit
= G m M/r^2 where M is mass of central body, in this case earth
so v^2 = GM /r
call moon orbit r = 1 and sun r = 372

so for moon 1.4^2 = G M/1
and sun v^2 = G M/372
v^2 = 1.42^2/372
v = 1.42/sqrt(372)

Answer 2

1.4mm/s=

.0014m/s(60sec/min)*60min/hr(24hr/day)(14day/halfMoonCycle)=1690m about a land mile.
Hmmmm. So I wonder in the hypothesis, what direction is the 1.4mm/s?

Answer 3

Direction? The Moon revolves the Earth counter-clockwise. The Sun is rotating counter-clockwise. But these are not related to the question as it does not ask for direction.

Answer 4

To calculate the rate at which the Sun is "falling," we need to consider the gravitational forces between the Sun, Earth, and Moon.

First, let's find the rate at which the Moon is falling around the Earth. You mentioned that the Moon is falling at a rate of 1.4 mm/s. This rate is determined by the gravitational force between the Earth and the Moon.

Now, let's use the given information that the Sun is 372 times farther from the Earth than the Moon. This means that the distance between the Earth and the Sun is 372 times the distance between the Earth and the Moon.

Since the gravitational force decreases with distance, we can assume that the gravitational force between the Sun and the Earth is 1/(372^2) times the force between the Earth and the Moon.

So, the rate at which the Sun is "falling" can be calculated by multiplying the rate at which the Moon is falling around the Earth (1.4 mm/s) by the ratio of the gravitational forces.

Let's calculate it:

Rate of Sun "falling" = Rate of Moon falling * (Force between Sun and Earth / Force between Earth and Moon)
= 1.4 mm/s * (1/(372^2))

Now, let's calculate it:

Rate of Sun "falling" ≈ 1.4 mm/s * 7.22 × 10^-6
≈ 1.008 × 10^-5 mm/s

So, the rate at which the Sun is "falling" is approximately 1.008 × 10^-5 mm/s, or 10.08 nanometers per second.

Therefore, the answer is approximately 1.008 × 10^-5 mm/s, which is equivalent to 10.08 nanometers per second.