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Brent
2006-Mar-02, 05:53 PM
Can anybody explaine to me how to find weight in newtons of something whos mass is measured in kilograms? Assume that the mass is known.

grant hutchison
2006-Mar-02, 06:06 PM
You need to know g, the acceleration due to gravity. On the Earth it's around 9.8 N/kg.
Multiply the mass by g, and you'll have the weight.

Grant Hutchison

Bob B.
2006-Mar-02, 06:06 PM
W = mg

where W is weight (N), m mass (kg), and g the acceleration of gravity. The standard value of g for Earth is 9.80665 m/s^2.

tony873004
2006-Mar-02, 07:23 PM
W = mg

where W is weight (N), m mass (kg), and g the acceleration of gravity. The standard value of g for Earth is 9.80665 m/s^2.
I've never seen it expressed to that many digits. I've always assumed that it's because from place to place it varies enough in the 0.00x position that expressing it further was meaningless unless you specified an exact location. Maybe that is a planet-wide average?

grant hutchison
2006-Mar-02, 08:04 PM
I've never seen it expressed to that many digits. I've always assumed that it's because from place to place it varies enough in the 0.00x position that expressing it further was meaningless unless you specified an exact location.The (rounded) first decimal digit is all you can express with certainty, in fact. According to the fourth edition of Allen's Astrophysical Quantities, the Earth's reference surface gravity varies from 9.78031 m/s² at the equator to 9.83217 m/s² at the poles.
9.80665m/s² is the exact definition of the non-standard unit of acceleration used in aeronautics, 1g. It's not quite equal to the latitudinal average, the areal average, or the acceleration at 45º latitude: I don't know why that exact figure was adopted.

Grant Hutchison

tony873004
2006-Mar-02, 10:31 PM
Thanks, Grant. ...Another 'g' question.

I see g commonly referred to as acceleration due to gravity. So at the equator, where your downward acceleration is GM/r^2 minus centrifugal force from Earth's rotation, does g simply represent the gravity part of it, or the whole thing. If it's the whole thing, is it correct to say that g is not the acceleration due to gravity, but rather it is the net downward force an object experiences at the surface of Earth?

grant hutchison
2006-Mar-02, 10:52 PM
Again from Allen's: "Gravity includes the gravitational attraction of the Earth's mass and the centrifugal acceleration of the Earth's rotation." So figures for Earth's surface gravity, usually symbolized g, include the centrifugal component.

Grant Hutchison

Ken G
2006-Mar-03, 11:28 AM
And note the centrifugal term at the equator scales with the square of the rotation speed. The speed needed to be in orbit is 8 km/s, so these two facts imply that the relative size of the centrifugal effect is the rotation speed, 0.5 km/s, divided by the orbital speed, 8 km/s, all squared. That's (0.5/8)^2 = 0.004 out of the full g. So we should be looking for a 0.04 m/s^2 shift in g at the equator due to the centrifugal effect. Grant's number is more like 0.05 m/s^2, so where does the last 0.01 m/s^2 come from? That must be the contribution due to the oblateness of the Earth.

tony873004
2006-Mar-03, 06:28 PM
And note the centrifugal term at the equator scales with the square of the rotation speed. The speed needed to be in orbit is 8 km/s, so these two facts imply that the relative size of the centrifugal effect is the rotation speed, 0.5 km/s, divided by the orbital speed, 8 km/s, all squared. That's (0.5/8)^2 = 0.004 out of the full g. So we should be looking for a 0.04 m/s^2 shift in g at the equator due to the centrifugal effect. Grant's number is more like 0.05 m/s^2, so where does the last 0.01 m/s^2 come from? That must be the contribution due to the oblateness of the Earth.

If you use the more accurate numbers of (4.638 / 7.901)^2 you get 0.003

I'm not sure where you got 0.005 from Grant's post, but I imagine you did:
1-(9.78031 / 9.83217) =0.005
Which is comparing the pole to the equator.

Comparing the "exact definition" to the equator value of g gives you
1-(9.78031 / 9.80665) = 0.003
which is your answer with more accurate inputs.

alainprice
2006-Mar-03, 10:21 PM
9.83217 - 9.78031 = 0.05186
The difference in acceleration at the surface of the earth varies by 0.05 m/s^2 between the equator and the poles.

Ken also pointed out that the centrifugal contribution(at the equator) to acceleration is 0.4%, or 0.004. If you take g(9.8 m/s2) and multiply by 0.004, the answer is roughly 0.04 m/s2.

tony873004
2006-Mar-04, 05:15 AM
Thanks. I read an extra 0 into it. That's why I couldn't see where Ken got that number.

Ken G
2006-Mar-04, 09:06 AM
This is interesting, if the 0.03 m/s^2 for the centrifugal acceleration at the equator is indeed more accurate, and I have no reason to doubt tony873004's numbers, then Grant's 0.05m/s^2 difference tells us the oblateness effect is 0.02 m/s^2 not 0.01 m/s^2. That would make it almost the same as the centrifugal effect. If so, this says that we launch rockets from Florida not just because of the rotation energy we pick up, but almost just as much from the "lift" out of the Earth's gravitational potential that we gain from the Earth's oblateness! That surprises me, but there are the numbers.