Earth's Velocity without the Moon: Formula & Answers

[YellowTaxi]

DH is correct..

Probably true tim_lou, but I don't think the effect would make a noticeable difference because we're dealing with basically a circular orbit, and I think the complexity only applies to elliptic orbits with extreme excentricity.

It's like arguing that Galileo was wrong because he ignored the effects of air viscosity when he dropped two masses off that tower

Pedantic

 

[tim_lou]

Re-reading the original question, the question is not precise enough to be answered, because the mechanism in which the moon disappeared is not specified, and/or some conditions (total angular momentum conserved?) are not specified.

And even if the moon ceases to exist, the Earth can stay in the sum's orbit with an arbitrary amount of angular momentum associated with spin. (the orbital angular momentum and spin angular momentum are decoupled in this situation).

 

[YellowTaxi]

what ?
So how much was I out by (1 second a year or maybe less)

Maybe we forgot to include the extra drag induced by new blades of grass growing near lake Baikal.

I'll go back and recalculate ...

 

[Janus]

Roughly,

earth to moon = 384 403 km
earth to sun = 150 000 000 km

Earth mass 5.9742 × 10^24 kilograms
Moon 7.36 × 10^22 kilograms

centre of mass is .'. = 384,403 * M/(M+E) from Earth
= 4678 km from centre of the Earth

Worst case is when the Earth is furthest from sun (moon closest) and r -> r + 4678 km (lol)
so where r = normally (on average) 150 000 000 km,
from Kepler #3,
T' = T * (r'/r)^3/2
= 365 * (150,004,678/150,000,000) ^3/2
=365.01707510616413248734767419157 days
(assumed a typical year is exactly 365 days)

so maximum difference is about 0.01707510616413248734767419157 of a day, per year.
= 24 minutes 35 seconds (not much)
minimum = 0 seconds

(p.s. I don't think tidal effects make any difference whatsoever, though I'm not absolutely certain of that.)

cheers, Yellow Taxi
good question

edit: corrected Earth to moon distance

You can't just plug 150,004,678,000 in for r' in the Kepler
formula. In this case r' is the semimajor axis of the new orbit and will be larger than the value you used.

Assume, for the sake of argument, that the original Earth-Moon orbit is circular. Now when the Earth is furthest from the Sun it is not only 4,678 km further from the Sun than the Earth-Moon barycenter, but it is also moving faster as it is also orbiting the barycenter at 12.5 m/sec. This means that it is traveling faster than it would for a circular orbit at that distance.

So if the Moon were to disappear at this moment, the Earth would enter a new elliptical orbit with its present distance at perhelion. This puts the new value of r' for this orbit at a further distance out.

Using 149,600,000km for r, the new r' works out to 149,735,000km

Plugging these values into the Kepler equation yields a difference in orbital periods of about 11.82 hrs; about 30 times the answer you got.