Moon's Escape Point: Beyond Earth's Gravity

[H8wm4m]

What is the distance from the moon where the gravitational influence of the Earth is weaker than that of the moon?
I haven't found it in any online encyclopedia and unfortunately Nasa.gov is not working for me. I would imagine it would be easy to find, but to my amazement I haven't even found a single mention of it.

 

[H8wm4m]

I read that the point where a body begins to fall towards the moon is nowhere near it should be, if the moon had the alleged one-sixth Earth gravity (0.1654 g). Apparently it should be in the 20,000 to 22,000 mile range (as in the older versions of Britannica), but after the apollo missions the accepted range changed to 40,000 to 38,000 mile range.

Soooo...
I decided to find out if this is true for myself, but I can't find it anywhere!
On top of that, I'm not sure exactly how to write an equation to detemine it either.

Any help on finding and/or determining it would be appreciated.
Thanks

 

[H8wm4m]

First I would use the inverse square law to detemine the distance from the Earth where gravity would accelerate a body at 1.622 m/s2 (sg of the moon)
I would then subtract this distance from the average distance between the Earth and moon, and divide the result by two.
I think this will give me the distance from the moon where acceleration would be equal on both sides.

There is probably a simpler way, but I don't know one.
So my question is:
Would this method work, and how would I put it into an equation?

P.s. forgive my terminology, I hope you all can understand what I am trying to get across

 

[Janus]

I read that the point where a body begins to fall towards the moon is nowhere near it should be, if the moon had the alleged one-sixth Earth gravity (0.1654 g). Apparently it should be in the 20,000 to 22,000 mile range (as in the older versions of Britannica), but after the apollo missions the accepted range changed to 40,000 to 38,000 mile range.

Soooo...
I decided to find out if this is true for myself, but I can't find it anywhere!
On top of that, I'm not sure exactly how to write an equation to detemine it either.

Any help on finding and/or determining it would be appreciated.
Thanks

Hmm, the distance from the moon where the gravitational force from the Earth and the moon exactly cancel is just about 1/10 the distance from the Earth to the moon or about 38,400 kilometers or 24,000 miles. Taking into account rounding out, I'd have to say that this accepted distance hasn't changed, rather the units of distance used to measure it are different between the two sources.

 

[tony873004]

<br /> \begin{array}{l}<br /> \frac{{GM_m }}{{d_m^2 }} = \frac{{GM_ \oplus }}{{d_ \oplus ^2 }} \\ <br /> \\ <br /> \frac{{M_m }}{{d_m^2 }} = \frac{{M_ \oplus }}{{d_ \oplus ^2 }}\,\, \\ <br /> \\ <br /> \,\,\frac{{M_m }}{{M_ \oplus }} = \frac{{d_m^2 }}{{d_ \oplus ^2 }}\, \\ <br /> \frac{{d_m }}{{d_ \oplus }} = \sqrt {\frac{{M_m }}{{M_ \oplus }}} \\ <br /> \\ <br /> d_m = d_ \oplus \sqrt {\frac{{M_m }}{{M_ \oplus }}} \\ <br /> \\ <br /> d_m = d_{m \oplus } - d_ \oplus \\ <br /> \\ <br /> d_ \oplus \sqrt {\frac{{M_m }}{{M_ \oplus }}} + d_ \oplus = d_{m \oplus } \\ <br /> \\ <br /> d_ \oplus \left( {\sqrt {\frac{{M_m }}{{M_ \oplus }}} + 1} \right) = d_{m \oplus } \\ <br /> \\ <br /> d_ \oplus = \frac{{d_{m \oplus } }}{{\left( {\sqrt {\frac{{M_m }}{{M_ \oplus }}} + 1} \right)}} \\ <br /> \\ <br /> d_ \oplus = \frac{{384000km}}{{\left( {\sqrt {\frac{{7.34 \times 10^{22} kg}}{{5.97 \times 10^{24} kg}}} + 1} \right)}} = 345671km \\ <br /> \\ <br /> d_m = 384000 - 345671 = 38329km \\ <br /> \end{array}<br />

<br /> \begin{array}{l}<br /> d_ \oplus = \frac{{384000km}}{{\left( {\sqrt {\frac{{7.34 \times 10^{22} kg}}{{5.97 \times 10^{24} kg}}} + 1} \right)}} = 345671km \\ <br /> \\ <br /> d_m = 384000 - 345671 = 38329km \\ <br /> \end{array}<br />

 

[Janus]

First I would use the inverse square law to detemine the distance from the Earth where gravity would accelerate a body at 1.622 m/s2 (sg of the moon)
I would then subtract this distance from the average distance between the Earth and moon, and divide the result by two.
I think this will give me the distance from the moon where acceleration would be equal on both sides.

This method will not give you the right answer. For one, you need to take into account the fact that the gravity of the moon also falls off by the inverse square law.

The method given by tony873004 is how you would do it.

 

[marcus]

Tony's method is correct but I would not use it if I actually wanted a quick estimate.

A good fact to remember is that the moon mass is 1/81
of the Earth mass.

the handbook value of the mass ratio is 81.3
but there's no need for such great precision, it would be OK to just remember that moonmass is about one EIGHTIETH of earthmass.

so you can almost do it in your head, or on back of envelope. Let R be the distance you want as a FRACTION of the total distance to moon.

Equal force in each direction, at that distance, says

\frac{1}{R^2} = \frac{1}{80} \frac{1}{(1-R)^2}

\frac{1 - R}{R} = \sqrt {1/80}

1/R = 1 + sqrt 1/80

R = \frac {1}{1 + \sqrt{1/80}} = 9/10

so the equal pull point is NINETY PERCENT of the way to the moon

the arithmetic is easy to do in your head because square root of 1/81 is 1/9
and 1 + 1/9 = 10/9
so when you invert that fraction you get 90%

=================

Mr H8...

You got confused by the SURFACE gravity. The ratio of surface gravity is ONE SIXTH
but the overall strength of gravity is determined by MASS (divided by distance squared)

MASS of moon is only 1/81 of earth---therefore at equal distance much much weaker.
If they had equal radii the moon surface gravity would be only 1/81 but part of that differnce is compensated
because the moon is SMALLER. So a person standing on surface is CLOSER to moon center---which makes him feel more.
So the way it works out is the surface gravity is 1/6

 

[Janus]

Tony's method is correct but I would not use it if I actually wanted a quick estimate.

A good fact to remember is that the moon mass is 1/81
of the Earth mass.

the handbook value of the mass ratio is 81.3
but there's no need for such great precision, it would be OK to just remember that moonmass is about one EIGHTIETH of earthmass.

so you can almost do it in your head, or on back of envelope. Let R be the distance you want as a FRACTION of the total distance to moon.

Equal force in each direction, at that distance, says

\frac{1}{R^2} = \frac{1}{80} \frac{1}{(1-R)^2}

\frac{1 - R}{R} = \sqrt {1/80}

1/R = 1 + sqrt 1/80

R = \frac {1}{1 + \sqrt{1/80}} = 9/10

so the equal pull point is NINETY PERCENT of the way to the moon

the arithmetic is easy to do in your head because square root of 1/81 is 1/9
and 1 + 1/9 = 10/9
so when you invert that fraction you get 90%

=================

This is pretty much how I arrived at my figure.

 

[D H]

All of you are correct in the sense that the gravitational acceleration toward the moon equals the acceleration toward the Earth at about 38,000 km or 23,000 miles. The OP (actually, his second post) refers to another figure, 38,000 miles. It could simply be a units problem that caused this confusion. I suspect a deeper issue: The Hill sphere. In a very real sense it is the Hill Sphere rather than the equipotential surface that decides which body predominates. Example: The Moon is inside the Earth's Hill Sphere but outside the Earth/Sun equipotential surface (the gravitational acceleration of the Moon toward the Sun is greater than toward the Earth). As we don't tolerate lunatics at Physics Forums (pun intended), I think we can all agree that the Moon is indeed in orbiti about the Earth.

Back to the Moon: The Moon's Hill Sphere is about 60,000 km / 38,000 miles.

 

[H8wm4m]

Thanks, that helps
Is nasa.gov working for anybody?
Any other site I visit works just fine

 

[chrisina]

H8,

the distance from the moon where the gravitational influence of the Earth is weaker than that of the moon is as explained and calculated by Marcus, Tony, Janus, 38,400 km.

The tricky part is that this is not the distance where an object will start gravitating around the moon.

Indeed, the calculation as shown above by Tony and Marcus only equates the gravitational forces of the moon and Earth (which was your original question), but as suggested by DH if you want to find out the gravitational sphere of influence of the moon, you need to take into acount in the calculation a 3rd force, which is the centrifugal force in a frame of reference rotating around the Earth at the angular frequency of the moon. This is what gives you the Hill sphere, which extends between two Lagrange points.


In order to make this calculation, you need to make use of Lagrangian mechanics and it is much more complicated than the calculation as shown above.

The result is that, the distance from the moon turns out to be approximately 61,500 km, as you can see larger than the 38,400 km mentionned above, because the centrifucal force works in favour of the moon. For example at 50,000 km from the moon, the gravitational force of the moon is weaker than that of the earth, but a body placed there with still orbit around the moon, which itself of course rotates around the earth.



so d = D x (1/ 244) ^(1/3)

 

[tony873004]

so d = D x (1/ 244) ^(1/3)

That's the Hill Sphere formula, where 1/244 is the ratio of the Moon's mass : 3x Earth's mass.

This is the maximum distance anything can possibly orbit the Moon. But it's worth noting that objects close to the Hill Sphere are generally not stable. Only an object orbiting the Moon in a direction retrograde to the Moon's motion around the Earth will be able to complete some orbits. Prograde objects are unstable until you get a bit deeper into the Hill Sphere. As a rough estimate, an object should be no more than about 1/3 of the distance to the edge of the Hill Sphere to maintain a stable, nearly-circular, prograde orbit.

It's also interesting to note that although Earth's Hill Sphere with respect to the Sun extends out to about 1.5 million kilometers, the presence of the Moon makes all prograde orbits external to the Moon unstable.

Here's a page I made of some useful astro calculators. Hill Sphere is in the middle of the page:
http://www.orbitsimulator.com/formulas/