Question from a child that I was unable to answer

[Nimby]

Eric Mc said:

What hasn't been explained?

If a long object is in orbit around a planetary body and one end is pointed directly at the centre of the planetary body with the other end pointed directly away, the long object will experience a stronger gravitational pull at the end nearest the planetary body and a proportionally weaker pull at the end furthest away from the planetary body.

The difference between the strength of the gravitational pull between either ends is what I described above as "the gravitational gradient".

That's the gist of Larry Niven's SF short story "Neutron Star"; the effect can be quite dramatic when the surface gravity is in the millions.

[Eric Mc]

That's why I mentioned a black hole earlier. Any object exerting a massive gravitational pull (such as a black hole or a neutron star) would stretch any object that ventured too close to it into a long thin line of molecules which would spiral in towards it.

Following on from NASA's tethered experiments with Gemini 11, they had another go at something similar on a couple of Space Shuttle flights. These didn't go to plan either - mainly because the unreeling mechanism of the tether jammed or because the tether itself snapped.

Edited by Eric Mc on Monday 28th November 16:17

[Hugo a Gogo]

SpeckledJim said:

Hugo a Gogo said:

that's different, using geostationary orbit to balance it, not really to do with reduced gravity

It's a similar principle though. If the vertical pin's length was over double that of it's geostationary orbit height, then it would be so 'weightless' at ground level that you would have to tether it to the ground to stop it taking-off.

That's not something that would happen to the same pin lying horizontally on the ground.

So it demonstrates the principle that 'heavy stuff' weighs more/less depending on its position relative to earth.

No it's not to do with reduced gravity though. For example it wouldn't work at the poles. It weighs less because of its motion relative to the earth. A brick is weightless when you chuck it up in the air. Chuck it fast enough and it orbits

[Hugo a Gogo]

Incidentally, the 'distance from centre of gravity' calculation doesn't work once you start going deep below the surface. If went to the centre of the earth you'd be weightless, because the gravity would equal from all sides

[Derek Smith]

Perhaps a little harsh of me. My apologies to anyone I've offended.

The lass knows about inverse square law. The whole thing started with a diagram of the Solar System stuck on the bedroom wall. This generated a query about tides, which she took on board. She asked for clarification of some things, such as the double bulge.

I was expecting to not only to be able to confirm what she was working towards herself, but come up with reasons, in the sense of:

Sticking with the needle. We have two: a horizontal one and a vertical one. Assuming that the centre of attraction is a single point, how far up the vertical one can the horizontal one go before the weights are equal?

In other words, is there a formula, a little algebraical bit, that can explain the constants sufficiently clearly for a 12 year old, albeit a rather sharp one?

[SpeckledJim]

Hugo a Gogo said:

SpeckledJim said:

Hugo a Gogo said:

that's different, using geostationary orbit to balance it, not really to do with reduced gravity

It's a similar principle though. If the vertical pin's length was over double that of it's geostationary orbit height, then it would be so 'weightless' at ground level that you would have to tether it to the ground to stop it taking-off.

That's not something that would happen to the same pin lying horizontally on the ground.

So it demonstrates the principle that 'heavy stuff' weighs more/less depending on its position relative to earth.

No it's not to do with reduced gravity though. For example it wouldn't work at the poles. It weighs less because of its motion relative to the earth. A brick is weightless when you chuck it up in the air. Chuck it fast enough and it orbits

True, it is not the same thing.

[Einion Yrth]

Derek Smith said:

In other words, is there a formula, a little algebraical bit, that can explain the constants sufficiently clearly for a 12 year old, albeit a rather sharp one?

F = GMm/R^2

For each and every "part" of the body concerned, split it however you find convenient and sum over the parts.


edited: missed rather important divide sign.

Edited by Einion Yrth on Monday 28th November 20:12

[RobM77]

Hugo a Gogo said:

Incidentally, the 'distance from centre of gravity' calculation doesn't work once you start going deep below the surface. If went to the centre of the earth you'd be weightless, because the gravity would equal from all sides

Yes, the principle does still work. The reason you'd be weightless at the centre of the earth is because at that point your distance from the centre of gravity is zero, not because the principle of the distance from the CofG doesn't work. The only difference is that outside the sphere it's an inverse square relationship between gravity and distance, but once you get inside a sphere you have the sphere's mass pulling you from all directions, which sums to make it a linear relationship.

[Hugo a Gogo]

What is the force of gravity one metre from the centre of the earth then?

[RobM77]

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

The linearity inside a sphere relationship assumes a consistent density within the sphere, which of course the earth doesn't have, and a perfect sphere, which of course the earth isn't. However, if it was a consistent density all the way through and a perfect sphere, then the force you felt 1 metre from the centre would obviously just be a fraction of the force you feel at the earth's surface. That fraction would be simply r/R, where r is your distance from the centre and R is the radius of this fake 'perfect' earth.

[anonymous-user]

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish

[RobM77]

wsurfa said:

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish

The earth's average radius is 6371km, so at 1 metre from the centre, your weight (i.e. the gravitational force pushing downwards due to your mass) would be 0.016% of what it is at the surface. This assumes a uniform density, which as we know is not the case. In actual fact, the earth gets denser as you go inwards, so the real figure would be slightly higher than this.

[anonymous-user]

RobM77 said:

wsurfa said:

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish

The earth's average radius is 6371km, so at 1 metre from the centre, your weight (i.e. the gravitational force pushing downwards due to your mass) would be 0.016% of what it is at the surface. This assumes a uniform density, which as we know is not the case. In actual fact, the earth gets denser as you go inwards, so the real figure would be slightly higher than this.

Using 12gcm3 I think its close to 0.03% (ish)

[RobM77]

wsurfa said:

RobM77 said:

wsurfa said:

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish

The earth's average radius is 6371km, so at 1 metre from the centre, your weight (i.e. the gravitational force pushing downwards due to your mass) would be 0.016% of what it is at the surface. This assumes a uniform density, which as we know is not the case. In actual fact, the earth gets denser as you go inwards, so the real figure would be slightly higher than this.

Using 12gcm3 I think its close to 0.03% (ish)

Thanks - I was about to work that out having looked up the density of the inner core!

ETA: I'd weigh just a smidge over 2kg

[Hugo a Gogo]

So not F = GMm/R^2 then?

[RobM77]

Hugo a Gogo said:

So not F = GMm/R^2 then?

That's Newton's universal law, yes. However, as I explained earlier, inside a sphere the gravity from each direction sums to make it a linear relationship.

[Hugo a Gogo]

Hugo a Gogo said:

Incidentally, the 'distance from centre of gravity' calculation doesn't work once you start going deep below the surface. If went to the centre of the earth you'd be weightless, because the gravity would equal from all sides

which is what I said

[Abbott]

Hmm interesting. If you were a free body at the centre of the earth then the gravitaional effecrt would be zero but would you not be subject to the gravitational pull of the moon/sun which would drag you away from the centre. If so by how much.
(Disclaimer: I could be talking B*****ks)

[V8LM]

RobM77 said:

wsurfa said:

RobM77 said:

wsurfa said:

Hugo a Gogo said:

What is the force of gravity one metre from the centre of the earth then?

probably about 0.01% ish of that acting on you at the surface

assuming you're 100kg then c 0.1N 1km from the centre vs c 1000n at the surface

ish - very much ish

The earth's average radius is 6371km, so at 1 metre from the centre, your weight (i.e. the gravitational force pushing downwards due to your mass) would be 0.016% of what it is at the surface. This assumes a uniform density, which as we know is not the case. In actual fact, the earth gets denser as you go inwards, so the real figure would be slightly higher than this.

Using 12gcm3 I think its close to 0.03% (ish)

Thanks - I was about to work that out having looked up the density of the inner core!

ETA: I'd weigh just a smidge over 2kg

Aren't we a few orders of magnitude out here? (radius is 6 371 000 m).

[SpeckledJim]

Abbott said:

Hmm interesting. If you were a free body at the centre of the earth then the gravitaional effecrt would be zero but would you not be subject to the gravitational pull of the moon/sun which would drag you away from the centre. If so by how much.
(Disclaimer: I could be talking B*****ks)

Yes, whether you are on earth, or inside earth, you and the moon will be attracted to each other by gravity.

The attraction between you and the sun may be balanced by your inertia as you travel in a circle around it? Not sure.