[Seany (OP)]

What about if the explosion caused the moon lose half its mass?

[lyner]

That 1/2 mass would have to go somewhere at some speed. The total momentum and angular momentum would still remain unchanged.

[Seany (OP)]

The orbit time would also remain unchanged?

[Make it Lady]

On the programme the explosion didn't seem to remove any mass. The moon was pushed out of orbit by some weird Newton's Third law of motion effect!!!

[lyner]

The orbit time would also remain unchanged?

the orbit time of what? The half that went off in one direction would have a different orbit and the other bits would have their own orbits. Earth wouldn't have to change significantly if the event was short lived. The reason I say that is because the force on the Earth would start off as the existing gravitational attractive force and then drop to zero at a huge separation distance. The change of momentum of Earth would be equal to this force times the time it acted for. Less time - less momentum change. For an explosive event, the change would be slight.
Imagine trying to drag a magnet across a table with another magnet; if you whipped your magnet away quickly, you couldn't budge it - you'd need to keep your magnet close for a long time and 'tease' the other magnet along. The Moon scenario would be the same sort of idea.

[Seany (OP)]

The orbit time would also remain unchanged?

the orbit time of what? The half that went off in one direction would have a different orbit and the other bits would have their own orbits. Earth wouldn't have to change significantly if the event was short lived. The reason I say that is because the force on the Earth would start off as the existing gravitational attractive force and then drop to zero at a huge separation distance. The change of momentum of Earth would be equal to this force times the time it acted for. Less time - less momentum change. For an explosive event, the change would be slight.
Imagine trying to drag a magnet across a table with another magnet; if you whipped your magnet away quickly, you couldn't budge it - you'd need to keep your magnet close for a long time and 'tease' the other magnet along. The Moon scenario would be the same sort of idea.

The half that flies away.. Forget about it.

The half that stays in orbit.. Would that one still orbit the earth once every 28 days?

[turnipsock]

Actually, this question isn't that daft. There have been a number of plans to explode an atomic bomb on the moon.

If the earth and moon orbit the sun as one mass, then we got rid of the moon, the mass would have decreased and the earth would then drift away from the sun.

I have a lot of pictures of the moon taken with the TSGT if anybody is interested. Its good to keep pictures to remind us what things looked like when they are gone.

Isn't the moon's mass an 1/8 of the earths?

[Seany (OP)]

If the earth and moon orbit the sun as one mass, then we got rid of the moon, the mass would have decreased and the earth would then drift away from the sun.

Yes yes yes! This is what I thought. But apparently even if the mass of the moon went down, it would find another orbit..

[Seany (OP)]

But why would there be plans to blow up the moon with an atomic bomb?

[turnipsock]

But why would there be plans to blow up the moon with an atomic bomb?

I don't think it was to blow it up but a show of stength during the cold war. Both the Americans and Russians had plans to do this.

http://en.wikipedia.org/wiki/Project_A119

http://www.svengrahn.pp.se/histind/E3/E3orig.htm

The americans idea was to produce a mushroom cloud that would be visable from earth!

[Seany (OP)]

Would it have actually made the Earth fall out of orbit of the Sun?

[lyner]

If the earth and moon orbit the sun as one mass, then we got rid of the moon, the mass would have decreased and the earth would then drift away from the sun.

That is just nonsense. The orbit diameter might change by a minuscule amount because of the change of position of the 'mutual centre of mass' of E,S &M. The most basic orbit theory tells you that the mass of a minor object in orbit is not relevant to its orbit time. What counts is its kinetic energy relative to its gravitational potential energy - mass is a common factor in each and is irrelevant. I could write down the sums but the web is full of this sort of elementary theory; mass just doesn't come into the orbit formula for small objects.
A simple argument goes like this - if you reducfe the mass then there is less gravitational force BUT the mass it acts on has decreased pro rata - no net effect.

[Seany (OP)]

Wouldn't a smaller mass be further away from the thing it's orbitting?

[lyner]

Why would it?
To go round in a circle of a certain diameter requires a certain amount of force per kilogram of the object. Each kilogram of the object experiences a pro-rata force keeping it in that orbit.
A pea doesn't require much force to keep it in a particular orbit but its attraction is a small force. A Centurion tank will need much more force to make it follow the same orbit but, because its mass is so great, it gets much more atttractive force. If the two are going at the same tangential speed, they will orbit at the same distance.
It all depends on where and at what speed they started off, if you like.
Just why they go in nearly circular orbits and why the orbits of all the planets are the way they are is a more complicated matter but let's start with the basics.

[Seany (OP)]

Because if a small mass.. Such as something that ways 100 tonnes. Which is pretty light for something like a meteorite or something. If that was to have an orbit with the sun, and was close to the sun.. The sun's gravitational pull would drag it straight in.

So for the small mass of block thing to orbit the sun, it must be far off, where the sun's gravitational pull is further away..

[lyner]

'fraid that's wrong still. Oh ye of little faith!
The effect on the motion of a mass depends upon the force and the value of the mass (Newton's second law of motion).
In the same way as orbits are independent of mass,  objects of low and high mass both fall to Earth at the same rate. (Well know experiment and you can more or less prove it for yourself if your light object is not so light that air resistance starts to have an effect - no feathers).
Here's chapter and verse ( you need to wait for A level before they do this at School for you. The Maths is reasonable, though, and is a clincher!!):
Centripetal force needed to keep an object in a circular path of radius r at speed v is
F(centripetal) = mv*v/r - like a conker on a string.

Force attracting it towards the Sun (Sun's Mass = M, G is the Gravitational Constant)
F(gravitational) = m*M*G/r*r

When the object is in a stable orbit, these two forces are equal / balanced, so the object won't move away or towards the Sun, just stay moving in a circle.
so
m*v*v/r = m*M*G/r*r
the m's cancel and so does one of the r's
so:
v*v = M*G/r
re-arrange it to find the radius of orbit
r=M*G/v*v
m doesn't come into it; all that counts is the speed. This assumes that M is a lot, lot bigger than m, which it is and is the simplest case of a circular orbit - but it applies for elliptical orbits too.

[Seany (OP)]

Way too complicated for my likings..

But I think I'm getting it. []

[turnipsock]

'fraid that's wrong still. Oh ye of little faith!
The effect on the motion of a mass depends upon the force and the value of the mass (Newton's second law of motion).
In the same way as orbits are independent of mass,  objects of low and high mass both fall to Earth at the same rate. (Well know experiment and you can more or less prove it for yourself if your light object is not so light that air resistance starts to have an effect - no feathers).
Here's chapter and verse ( you need to wait for A level before they do this at School for you. The Maths is reasonable, though, and is a clincher!!):
Centripetal force needed to keep an object in a circular path of radius r at speed v is
F(centripetal) = mv*v/r - like a conker on a string.

Force attracting it towards the Sun (Sun's Mass = M, G is the Gravitational Constant)
F(gravitational) = m*M*G/r*r

When the object is in a stable orbit, these two forces are equal / balanced, so the object won't move away or towards the Sun, just stay moving in a circle.
so
m*v*v/r = m*M*G/r*r
the m's cancel and so does one of the r's
so:
v*v = M*G/r
re-arrange it to find the radius of orbit
r=M*G/v*v
m doesn't come into it; all that counts is the speed. This assumes that M is a lot, lot bigger than m, which it is and is the simplest case of a circular orbit - but it applies for elliptical orbits too.

so the orbit of the earth is really about velocity? Increase the velocity and the earth drifts outwards to new orbit, decrease the velocity and the earth goes into a lower orbit. (I'm not sure 'velocity' is the correct term, 'speed' may be better)

[ukmicky]

'fraid that's wrong still. Oh ye of little faith!
The effect on the motion of a mass depends upon the force and the value of the mass (Newton's second law of motion).
In the same way as orbits are independent of mass,  objects of low and high mass both fall to Earth at the same rate. (Well know experiment and you can more or less prove it for yourself if your light object is not so light that air resistance starts to have an effect - no feathers).
Here's chapter and verse ( you need to wait for A level before they do this at School for you. The Maths is reasonable, though, and is a clincher!!):
Centripetal force needed to keep an object in a circular path of radius r at speed v is
F(centripetal) = mv*v/r - like a conker on a string.

Force attracting it towards the Sun (Sun's Mass = M, G is the Gravitational Constant)
F(gravitational) = m*M*G/r*r

When the object is in a stable orbit, these two forces are equal / balanced, so the object won't move away or towards the Sun, just stay moving in a circle.
so
m*v*v/r = m*M*G/r*r
the m's cancel and so does one of the r's
so:
v*v = M*G/r
re-arrange it to find the radius of orbit
r=M*G/v*v
m doesn't come into it; all that counts is the speed. This assumes that M is a lot, lot bigger than m, which it is and is the simplest case of a circular orbit - but it applies for elliptical orbits too.

so the orbit of the earth is really about velocity? Increase the velocity and the earth drifts outwards to new orbit, decrease the velocity and the earth goes into a lower orbit. (I'm not sure 'velocity' is the correct term, 'speed' may be better)

Angular momentum is the word or words  your looking for

[lyner]

(I'm not sure 'velocity' is the correct term, 'speed' may be better)

I think I do mean velocity - I should have said tangential velocity. Then the attractive force produces an acceleration towards the centre - no change of speed but a change in velocity (i.e. direction).

Angular momentum is the word or words  your looking for
   

Except that introduces a new concept and a new formula and I was trying to keep it simple. I don't think the misconception would have arisen if seany was very familiar with angular momentum.