Newton's gravity equation

[ivylove]

Newton's gravity equation is represented as:

 

 

 

F = G m1 m2 / r^2

 

 

 

Cavendish experiment is used to justify Newton's gravity equation by calculating the value of Newton's constant G but Cavendish's experimental apparatus uses two lead spheres m1 = .73 kg and m2 = 158 kg separated by the distance of .23 meters that produces a force of 1.74 x 10^-7 N or 2 micrograms yet the measurement uncertainty in 1787 was one milligram. Can anyone explain what is going on?

Edited July 20, 2017 by ivylove

[swansont]

Can you provide a link to where you found this?

[ivylove]

check wiki under "Cavendish" and look for 2 micrograms or the force.

Edited July 20, 2017 by ivylove

[Strange]

check wiki under "Cavendish" and look for 2 micrograms or the force.

 

 

Searching for Cavendish on Wikipedia (which is what I assume you mean) brings up several possible pages. Only one of which looks like it might be relevant: https://en.wikipedia.org/wiki/Henry_Cavendish

 

The words "2 micrograms" do not appear, and there is only a passing reference to the force of gravity.

 

Perhaps you could be a little more specific than "here are some numbers I read somewhere or made up".

OK. I guess this is what you are referring to:

 

 

Cavendish's equipment was remarkably sensitive for its time.^[9] The force involved in twisting the torsion balance was very small, 1.74×10⁻⁷ N,^[12] about  ¹⁄_50,000,000 of the weight of the small balls.^[13] To prevent air currents and temperature changes from interfering with the measurements, Cavendish placed the entire apparatus in a wooden box about 2 feet (0.61 m) thick, 10 feet (3.0 m) tall, and 10 feet (3.0 m) wide, all in a closed shed on his estate. Through two holes in the walls of the shed, Cavendish used telescopes to observe the movement of the torsion balance's horizontal rod. The motion of the rod was only about 0.16 inches (4.1 mm).^[14] Cavendish was able to measure this small deflection to an accuracy of better than one hundredth of an inch using vernier scales on the ends of the rod.^[15]

https://en.wikipedia.org/wiki/Cavendish_experiment#The_experiment

 

So that summarises how the necessary accuracy was achieved. What is your question?

 

And where does your value of the measurement uncertainty being "one milligram" come from? Is that what you think the likely error in the mass of the balls was? Which would be an error of 1 part in 158,000 or 0.0006%. That sounds impressive. Do you have a source for this?

[studiot]

Newton's gravity equation is represented as:

 

 

 

F = G m1 m2 / r^2

 

 

 

Cavendish experiment is used to justify Newton's gravity equation by calculating the value of Newton's constant G but Cavendish's experimental apparatus uses two lead spheres m1 = .73 kg and m2 = 158 kg separated by the distance of .23 meters that produces a force of 1.74 x 10^-7 N or 2 micrograms yet the measurement uncertainty in 1787 was one milligram. Can anyone explain what is going on?

 

 

This is a good opportunity for you to understand the experimental method.

 

I think by 'measurement uncertainty' you mean 'measurement resolution'.

They are different.

 

So the actual force was very small. Did Cavendish directly observe it?

Of course not. He measured something else entirely.

This something else was the twisting of a 6 foot beam, against a distance scale.

So the effect of the small force was multiplied by a long lever arm.

 

Now another of Newton's Laws asserts that, no matter how small the net force, it will accelerate a body and continue to do so for as long as the net force is obtained.

 

So large effects can arise from small net forces if you wait long enough, which Cavendish did.

Note the natural period of swing was about '20 minutes'.

 

So back to the experimental method.

 

Cavendish did not directly observe the force.

G and the force were calculated from the data, but not by the formula you quote but by this one (modern version)

[math]G = \frac{{\tau \theta {d^2}}}{{Mml}}[/math]

 

Where

tau is a previously determined constant of suspension

Theta is the deflection angle

d is the distance, that you mentioned, between the big and small lead balls

M and m are the masses of the lead balls

l is the length of the support rod.

 

Now what you need to do, when discussing the experiment is to take that equation and develop the effect of measurement resolutions in each of those quantities, in accordance with the theory of errors.

 

Further in experimental methods it is often known that certain effects change the observed readings and so must be allowed for.

 

In Cavendish's case corrections were applied for:-

 

1) Each large Sphere also attracted the more distant of the small spheres

 

2) The attraction of the beam slightly icreases the deflecting couple.

 

3) The rods supporting the masses also cause attraction.

 

A further experimental technique to increase precision is to take multiple readings and generate some kind of average.

Cavendish took 29 readings for his average, but I am not sure of the statistical methods available to him for this calculation.

Edited July 21, 2017 by studiot

[swansont]

check wiki under "Cavendish" and look for 2 micrograms or the force.

 

 

That's not how this works. You provide the link. Anyone posting is expected to provide support for their arguments. "Go Google it" (or similar) does not suffice.

 

(and, as noted, your search parameter gets no hits on the wikipedia page for the Cavendish experiment)

[Strange]

So the actual force was very small. Did Cavendish directly observe it?

Of course not. He measured something else entirely.

This something else was the twisting of a 6 foot beam, against a distance scale.

So the effect of the small force was multiplied by a long lever arm.

 

 

Excellent post.

[studiot]

Thanks Strange, and +1 for finding the correct link, which says that the deflection distance was 0.16 inches, measured to .01 inches with vernier.

 

This seems reasonable as it implies a scale marked on divisions of 0.1 inches divided a further 10 by the vernier.

 

Now it is interesting what this means for the resolution of angle.

 

At a radius of 3 feet or 36 inches 0.01 inches the angle subtended is given by

 

[math]\frac{{\delta \left( \theta \right)}}{{360*60}} = \frac{{.01}}{{2\pi *36}}[/math]

 

Which is approximately 1 minute of arc. (Don't you just love it when things cancel out so neatly)

 

[math]\delta \left( \theta \right) = \frac{{.01*360*60}}{{2\pi *36}} = \frac{6}{{2\pi }} \approx 1[/math]

[ivylove]

Also, Newton's gravity equation is applied to an astronaut that has a mass of 50 kg in the space station that is 249 miles (400,727 m) from the surface of the earth (r = 6.371 x 10^6 m + .4 x 10^6 m = 6.771 x 10^6 m) forming a gravitational force using Newton's equation of:

 

 

F = G m1 m2/r^2 = (.6 x 10^-11) (50) (6 x 10^24) / (6.771 x 10^6) = 438.4 N or 44.7 kg

 

 

According to Newton's gravity equation, a 50 ky astronaut forms a 44.7 kg force that is pointing in the direction of the earth yet an astronaut in the international space station is weightless which proves Newton gravity equation is physically invalid. Can anyone see this?

Edited July 21, 2017 by ivylove

[Strange]

According to Newton's gravity equation, a 50 ky astronaut forms a 44.7 kg force that is pointing in the direction of the earth yet an astronaut in the international space station is weightless which proves Newton gravity equation is physically invalid. Can anyone see this?

 

 

Really? I mean, REALLY?

 

Did you leave school at 11, or something? Did you never study any science at all? How old are you? How have you managed to remains so unbelievably ignorant? I am stunned, shocked and amazed. Do you need someone to remind you to keep breathing in and out?

 

Have you never heard of "inertia"?

 

Have you never heard of an "orbit"?

 

http://www.astronautix.com/n/newtonsorbitalcannon.html

 

I can't believe that someone as monumentally ignorant and uninformed keeps posting threads trying to show that basic physics is wrong WHEN YOU DON'T HAVE A F*ING CLUE.

 

Jesus H Christ.

 

I think the mods should close this now and just ban you to stop you wasting any more time (yours as much as anyone else's).

Edited July 21, 2017 by Strange

[beecee]

 

 

Really? I mean, REALLY?

 

Did you leave school at 11, or something? Did you never study any science at all? How old are you? How have you managed to remains so unbelievably ignorant? I am stunned, shocked and amazed. Do you need someone to remind you to keep breathing in and out?

 

Have you never heard of "inertia"?

 

Have you never heard of an "orbit"?

 

http://www.astronautix.com/n/newtonsorbitalcannon.html

 

I can't believe that someone as monumentally ignorant and uninformed keeps posting threads trying to show that basic physics is wrong WHEN YOU DON'T HAVE A F*ING CLUE.

 

Jesus H Christ.

 

I think the mods should close this now and just ban you to stop you wasting any more time (yours as much as anyone else's).

Couldn't agree more! Obviously this poster has come here with an agenda.

Just one question though....What does the "H" stand for in Jesus H Christ? I always thought it was Jesus F Christ.

[Strange]

Just one question though....What does the "H" stand for in Jesus H Christ?

 

 

Hitler? Maybe. I actually have no hecking idea!

[Endy0816]

Wiki has everything

 

https://en.wikipedia.org/wiki/Jesus_H._Christ

[Sensei]

F = G m1 m2/r^2 = (.6 x 10^-11) (50) (6 x 10^24) / (6.771 x 10^6) = 438.4 N or 44.7 kg

If G = 6.67*10^-11 N*m^2/kg^2

Then why did you use .6*10^-11... ?

I have no idea...

[swansont]

Also, Newton's gravity equation is applied to an astronaut that has a mass of 50 kg in the space station that is 249 miles (400,727 m) from the surface of the earth (r = 6.371 x 10^6 m + .4 x 10^6 m = 6.771 x 10^6 m) forming a gravitational force using Newton's equation of:

 

 

F = G m1 m2/r^2 = (.6 x 10^-11) (50) (6 x 10^24) / (6.771 x 10^6) = 438.4 N or 44.7 kg

 

 

According to Newton's gravity equation, a 50 ky astronaut forms a 44.7 kg force that is pointing in the direction of the earth yet an astronaut in the international space station is weightless which proves Newton gravity equation is physically invalid. Can anyone see this?

Study up on centripetal force and get back to us.

[studiot]

Study up on centripetal force and get back to us.

 

 

Since the last thread was closed for a total lack of engagement with those offering scientific discourse, I rather hoped you would do the same here following the same action.

[swansont]

Since the last thread was closed for a total lack of engagement with those offering scientific discourse, I rather hoped you would do the same here following the same action.

We're almost there.

[Janus]

On 7/21/2017 at 1:49 PM, ivylove said:

According to Newton's gravity equation, a 50 ky astronaut forms a 44.7 kg force that is pointing in the direction of the earth yet an astronaut in the international space station is weightless which proves Newton gravity equation is physically invalid. Can anyone see this?

I'm not sure if the OP is still around to see this, or if he'll even bother to read and try to understand it, but:

Have you ever ridden in a fast moving car that reaches a dip in the road, and you get the "leaving your stomach behind" feeling?  This is just a lesser magnitude example as to what is happening for the astronauts.  At that moment, you are just slightly "lighter" with respect to the car than you are normally, and this is what your stomach is reacting to. If you where to be traveling so fast that the car's wheels left the road for some time, you would feel "weightless" and objects would "float" around in the car.    This is because your "weight" is a function of gravity pulling down on you and the car, but the ground pushes back on the car and through it to you.  When the car leaves the ground, both you and the car are still subject to the same force of gravity as before, but are both free to respond to it equally and are both in free fall. There is no force pushing up on you to resist gravity and you are "weightless".

The "Vomit Comet" which is a plane designed to simulate this effect for longer than you could achieve in the car, does it by flying in an arc that follows such a free-fall path. Again, while following this arc, you, and object in the plane, float around in a weightless condition, even though the force of gravity on you has not let up one bit.

The ISS is in a state of continuous free-fall.    While near the surface of the Earth and at normal speeds, such a free fall path will always end up intersecting with the ground eventually, at the altitude and speed of the ISS, as it curves towards the ground, due to the spherical shape of the Earth, the ground curves away also.   It basically end up in a state of constantly falling towards the Earth but it keeps missing it.  Since both the ISS and astronauts are following the same free fall path, the astronauts end up in the same weightless condition as in the car in the air or the Vomit Comet during the arc, even though the force of gravity acting on them is not much weaker than it would be at the surface.

[ivylove]

In replay to Post 3:30 Firday---------If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the motion would be constant yet if one were to accelerate the said astronaut toward Jupiter or Uranus, to five mph, you would find that the astronaut would in fact propagate towards either stellar bodies. Moreover, if it was the centripetal force that was causing the weightlessness it would be extremely difficult to move in the opposite direction as the direction of the centripetal motion since the centripetal force would form a particular equilibrium that would be altered by the described astronaut propagating in the direction that opposes the centripetal velocity yet

Kathleen Rubins

Astronaut

is shown moving in random directions in the international space station and not sticking to the sides of the space stationary when propagating in the opposite direction of the centripetal motion. Also, I read another post where the Newton's constant was being described but I must inform you that it was a typographic mistake and should read 6.67 x 10^11 which I believe if the reader would have worked backward would have certainly deduce using inductive reasoning that the describe value was a typographic error not an error of judgement.  Also, there is a post regarding the measurement of Cavendish's experiment where a 2 microgram mass can be measured using a long pole  which I certainly cannot understand. Would you explain this extremely important scientific discovery.  I will be waiting in utter anticipation waiting for your paper or any link to this magnificent achievement that has up until now been concealed.

 

"The ISS is in a state of continuous free-fall."  

 

What is This? I this statement a joke? (since someone brought up the centripetal force)

 

Edited July 26, 2017 by ivylove

[beecee]

13 minutes ago, ivylove said:

"The ISS is in a state of continuous free-fall."  

What is This? I this statement a joke?

Why would you say that? Ignorance of the fact that any orbital body is actually in free fall perhaps?

 

[beecee]

Here's a more detailed answer.......

https://www.quora.com/Is-orbital-free-fall-what-we-normally-mean-by-falling

"It's falling in the sense that the only force acting on you is gravity. As the equivalence principle puts forward, there is no way to tell (locally) whether you are falling or floating freely in space under such conditions (that is, the only force acting on you is gravity).

As you correctly point out, it's not falling in the sense of "getting closer to the Earth", but there is an intuitive way to think about this. When we typically "fall" we're just going straight down---the distance we move is precisely how much closer to the surface of the Earth we get. But we could also imagine someone falling who also is moving sideways. The distance she moves is a combination of her sideways motion and her downwards motion. 

Once we get far up enough and fast enough (sideways) we run into a problem: the Earth isn't flat! As we fall "down" the Earth curves away beneath us! (This is because we are moving sideways---to a place where the Earth is curving in this fashion). So being in orbit is just a delicate balance of moving fast enough sideways that the distance we're falling downward is exactly compensated by the distance the Earth curves away from us."

Edited July 26, 2017 by beecee

[Janus]

5 hours ago, ivylove said:

"The ISS is in a state of continuous free-fall."  

 

What is This? I this statement a joke? (since someone brought up the centripetal force)

 

No joke. Look up "Newton's Cannon".   For an object in orbit, gravity is the centripetal force that prevents it from flying off in a straight line.

Edited July 26, 2017 by Janus

[Eise]

Imagine 2 satellites orbiting the earth very close to each other. Do you have any problem with that? (You need Newton's laws to understand this.)

Now imagine that one satellite is bigger than the other, but both still orbiting very close to the other. Do you have any problem with that? (You need Newton's laws to understand this.)

Now assume the biggest one is hollow, and the smaller one is completely inside the other. Both are doing their own orbit, which happens to be the same. Do you have any problem with that? (You need Newton's laws to understand this.)

So the smaller satellite is 'weightless' in relation to the big satellite: it does not bounce any wall of the bigger satellite. Do you have any problem with that? (You need Newton's laws to understand this.)

Now suppose the bigger satellite is filled with air, and the smaller one is an astronaut... Do you have any problem with that? (You don't need Newton's laws to understand this.)

Edited July 26, 2017 by Eise

[Strange]

9 hours ago, ivylove said:

Also, there is a post regarding the measurement of Cavendish's experiment where a 2 microgram mass can be measured using a long pole  which I certainly cannot understand. 

Somehow your inability to understand a brilliantly clear exposition of very simple principle does not surprise me. 

Has it occurred to you that the problem may not be with all of physics (and the many thousands of brilliant people who have contributed to our understanding of the world). The problem just could be that you are not quite as clever as you think you are. 

Basically, if you personally do not understand something, it does not mean it is wrong.

Edited July 26, 2017 by Strange

[swansont]

8 hours ago, ivylove said:

If it was the centripetal force that was causing the weightlessness of the 50 kg astronaut then the motion would be constant

No, the motion would be circular.