Question

[Johnny5]

and frankly' date=' it does nt change very much does it?

[/quote']

 

Imperceptible changes, so no it doesn't change very much.

 

Mean: 9.8066 m/s^2

 

Equatorial g: 9.78 m/s^2

 

No not much.

[Johnny5]

I cant say that is important' date=' just a little thing to be ware of [/quote']

 

 

Oh I know it's important, I'm just not sure where it fits in, and you say I should beware of it, as if it does matter somehow, into a master cosmological model or something.

[Johnny5]

So a kilogram mass is going to weigh different amounts as you move it around----sometimes it weighs 9.80 newtons of force

sometimes it weighs 9.78 newtons of force

 

 

 

Yes, you have one kilogram of inertia, a solid steel object, and you measure its weight all over the earth, at different locations. Your scale is kind of like a voltmeter, and you are measuring the potential difference.

 

What is the primary variable responsible for the differences in gee force?

 

R, the distance to the center of mass of earth?

[Martin]

I will make it easy by telling you the gee acceleration at that altitude

 

At sealevel with R = 6400 km the gravity is 9.8 m/s^2

 

so the new R is 7000

so we made R larger by ratio 70/64

so we make gravity weaker by the square of that ratio

 

9.8 x (64/70)^2 = 8.2 meters per second squared

 

 

Look J5! to find the speed of the satellite in meters per second all you need to do is set V^2/R equal to 8.2.

[Johnny5]

PROBLEM

what speed does a satellite have to go if it is going in a circle orbit around the earth with radius 7000 km?

 

This is simple enough.

[Johnny5]

I will make it easy by telling you the gee acceleration at that altitude

 

At sealevel with R = 6400 km the gravity is 9.8 m/s^2

 

so the new R is 7000

so we made R larger by ratio 70/64

so we make gravity weaker by the square of that ratio

 

9.8 x (64/70)^2 = 8.2 meters per second squared

 

 

Look J5! to find the speed of the satellite in meters per second all you need to do is set V^2/R equal to 8.2.

 

Gravity weaker by the square of that ratio...

 

[math] \frac{v^2}{7000} = 8.2 [/math]

 

[math] v^2 = 8.2 (7' date='000,000) = 57,400,000

 

[/math']

 

 

 

[math] v = 7576 m/s [/math]

[Martin]

Oh I know it's important, I'm just not sure where it fits in, and you say I should beware of it, as if it does matter somehow, into a master cosmological model or something.

 

no just be aware of v^2/R

dont have to "beware"

 

not a part of any bigdeal cosmological model that I know of

 

it is more like a tool that gets used so often (like screwdriver or visegrip pliers everybody has in the bottom drawer of the kitchen cabinet)

[Martin]

Gravity weaker by the square of that ratio...

 

[math] \frac{v^2}{7' date='000,000} = 8.2 [/math']

 

great going. you will have it in another minute!

 

yes, v = 7.5 km/sec

that is the typical speed in low earth orbit

[Johnny5]

great going. you will have it in another minute!

 

but hey' date=' are we working in meters or kilometers?

 

dont you mean 7000000?[/quote']

 

 

We are working in SI, so meters, seconds, kilograms.

 

V = 7576 meters per second = 7.5 km/s

 

1000 m = 1 km

 

Wow you are going fast. I wasn't using a calculator before, now I have one, need it to keep up with you.

 

At sea level R= 6400 km g = 9.8 m/s^2

 

Then at 7000 km, acceleration of earths gravity has diminished. You have this...

 

At R=7000 km above center of earths inertia g = 8.2

 

[math] 6400 \frac{70}{64} = 7000 [/math]

 

WE make gravity weaker by the square of that ratio...

 

Force on test mass m at surface of earth given by:

 

[math] \vec F = \frac{GMm}{6400^2} [/math]

 

Force on same mass at higher altitude R=7000km given by:

 

[math] \vec F = \frac{GMm}{7000^2} = \frac{GMm}{(6400*70/64)^2} [/math]

[Martin]

it is a handy thing to remember that the earth is always going about one tenthousandth of the speed of light in its roughly circular orbit around sun

 

 

(it gives a convenient handle on the doppler effect of the earth's own orbit speed)

 

 

what is the acceleration due to the sun's gravity at this distance where we find ourselves

 

 

===========

hmmmm looks like I have to go out for the morning

 

well the earth is going 30 km/sec and it is some 150 million km from sun

so it should be possible to find our sunwards acceleration that is keeping us in a roughly circular orbit. Is this accelerations greater or less than the 9.8 earthwards acceleration we all feel standing here in our shoes?

[Johnny5]

it is a handy thing to remember that the earth is always going about one tenthousandth of the speed of light in its roughly circular orbit around sun

 

 

(it gives a convenient handle on the doppler effect of the earth's own orbit speed)

 

 

what is the acceleration due to the sun's gravity at this distance where we find ourselves

 

I can answer all of your questions, but you have to slow down, I can barely keep up. I want to answer them all.

[Johnny5]

i

===========

hmmmm looks like I have to go out for the morning

 

 

 

 

Ok, take care I have to go too, I will answer all the questions later.

 

Regards

 

[Johnny5]

To answer all your questions...

 

Force on test mass m at surface of earth given by:

 

[math] \vec F = \frac{GMm}{6400^2} [/math]

 

Force on same mass at higher altitude R=7000km given by:

 

[math] \vec F = \frac{GMm}{7000^2} = \frac{GMm}{(6400*70/64)^2} [/math]

 

So

 

[math] \vec F = \frac{GMm}{6400}(64/70)^2 = 8.192 m/s^2[/math]

 

Let X = (64/70)²

 

So, the force on the test mass m at the surface of the earh, is reduced by X when that same mass is raised another 1000km.

 

At the surface of the earth, the weight of the object is mg, where g is the acceleration of gravity at the surface of the earth. As long as we can equate the force of gravity with weight, we have:

 

 

 

[math] W_{6400}= mg_{6400} = \frac{GMm}{6400^2} [/math]

 

Now, your weight at 7000 km above earths center of gravity would be given by:

 

[math] W_{7000}= mg_{7000} = \frac{GMm}{6400}X^2 [/math]

 

So...

 

[math] W_{7000}= mg_{7000} = mg_{6400} X^2 [/math]

 

So we can draw the following inference:

 

[math] g_{7000} = g_{6400} X^2 [/math]

 

[math] X = (\frac{64}{70})^2 = .836 [/math]

 

So...

 

[math] g_{7000} = (9.8 m/s^2) (.836) [/math]

 

I will make it easy by telling you the gee acceleration at that altitude

 

At sealevel with R = 6400 km the gravity is 9.8 m/s^2

 

so the new R is 7000

so we made R larger by ratio 70/64

so we make gravity weaker by the square of that ratio

 

9.8 x (64/70)^2 = 8.2 meters per second squared

 

 

Look J5! to find the speed of the satellite in meters per second all you need to do is set V^2/R equal to 8.2.

 

Ok' date=' now I understand what you meant when you said, we made gravity weaker by the square of that ratio. You could also have said, we decreased the acceleration due to gravity, by a factor equal to the square of that ratio. Yes, it all works out.

 

You are at some field point (x,y,z) which is a distance R away from the source of the field. If you release a test mass m at that field point, it will accelerate towards the center of gravity (the source) at a rate g[sub']R[/sub].

 

Now raise the test mass m higher, and you will decrease the force of gravity on m by some unknown factor X.

 

The force on m at distance R1 from the center of inertia is given by:

 

[math] F1 = m\frac{GM}{R_1^2} [/math]

 

The force on m at a distance R2 from the center of inertia is given by:

 

[math] F2 = m\frac{GM}{R_2^2} [/math]

 

Let R2 be further away from the center of gravity than R1. The ratio of F2 to F1 is:

 

[math] \frac{F2}{F1} = (\frac{R_1}{R_2})^2 [/math]

 

Now, notice that the weight of the test mass m at R1 is given by:

 

[math] W_1 = mg_{R1} [/math]

 

And the weight of the test mass m at R2 is given by:

 

[math] W_2 = mg_{R2} [/math]

 

But W2 = F2, and W1=F1 so that we have:

 

[math] F_1 = m g_{R1} [/math]

[math] F_2 = m g_{R2} [/math]

 

From which it follows that:

 

[math] \frac{F2}{F1} = \frac{g_{R2} }{g_{R1} } [/math]

 

Therefore:

 

[math] \frac{g_{R2} }{g_{R1} } = (\frac{R_1}{R_2})^2 [/math]

 

Which result you used. QED

[Johnny5]

it is a handy thing to remember that the earth is always going about one tenthousandth of the speed of light in its roughly circular orbit around sun

 

Describe the frame this is true in.

 

 

(it gives a convenient handle on the doppler effect of the earth's own orbit speed)

 

 

How so?

 

what is the acceleration due to the sun's gravity at this distance where we find ourselves...

 

V = 30 km/sec a

R = 150 million km from sun

 

...so it should be possible to find our sunwards acceleration that is keeping us in a roughly circular orbit. Is this accelerations greater or less than the 9.8 earthwards acceleration we all feel standing here in our shoes?

 

 

 

We can solve this like we solved the problem on earth.

 

[math] \vec a_c = \frac{v^2}{R} = \frac{30^2}{150 \times 10^6} [/math]

 

[math] \vec a_c = \frac{900}{150 \times 1' date='000,000} =\frac{9}{150 \times 10,000} [/math']

 

[math] \vec a_c =\frac{3}{5 \times 100,000} = \frac{3}{5} \times 10^{-5} = .6 \times 10^{-5} [/math]

 

 

[math] \vec a_c = 6 \times 10^{-6} [/math]

 

And the units are km/s^2.

 

1000 m = 1 km

 

so...

 

 

[math] \vec a_c = 6 \times 10^{-3} m/s^2[/math]

 

So a_c = .006 meters per second squared. This is far less than 9.8 m/s^2.

[swansont]

Assuming that he is in free-fall... As he falls' date=' the earth rotates a little so that he wont hit the surface of the earth right above where you drop him at. The earth will rotate slightly underneath him as he falls. Suppose you drop him at a height of 9.78 meters above the surface of the earth. He will strike the ground 1 second later, but in this time, the point on the surface of the earth which he was dropped above, will move slightly. It will accelerate at .0339 m/s^2 underneath him.

[/quote']

 

But if the person is stationary with respect to the ground before the freefall, that means he is moving with the same angular speed.

[J.C.MacSwell]

But if the person is stationary with respect to the ground before the freefall, that means he is moving with the same angular speed.

 

There is an effect. But I think it's smaller and in the opposite direction. (than what J5 was describing)

[J.C.MacSwell]

Assuming that he is in free-fall... As he falls' date=' the earth rotates a little so that he wont hit the surface of the earth right above where you drop him at. The earth will rotate slightly underneath him as he falls. Suppose you drop him at a height of 9.78 meters above the surface of the earth. [b']He will strike the ground 1 second later[/b], but in this time, the point on the surface of the earth which he was dropped above, will move slightly. It will accelerate at .0339 m/s^2 underneath him.

.

 

1 second later he will be travelling at 9.78 m/s (in the downwards direction) but will have dropped only 4.89 metres. His horizontal/rotational velocity/vector should exceed that of the Earth.

[J.C.MacSwell]

As to picking a spot on the moon the spin effect is obviously small due to the 27.3 day rotation time. The tidal effect is also small but I think it would be tens or hundreds of times larger in this case.

 

Sorry if I'm guessing while you guys are obviously doing the work. I've enjoyed following this thread and thought I might at least stick my neck out on an educated (?) guess.

[Martin]

...

so...

 

 

[math] \vec a_c = 6 \times 10^{-3} m/s^2[/math]

 

So a_c = .006 meters per second squared. This is far less than 9.8 m/s^2.

 

I like that answer, personally. I think it is right. Anybody want to double check what J5 says?

 

I think that the acceleration towards the sun, due to the sun's gravity, at our present distance away from it, is 0.006 meters per second square like what he says. Would be nice if someone else would go thru the arithmetic though, to catch any careless error.

[Martin]

BTW J5, jupiter like the earth has a roughly circular orbit and is 5 times farther from the sun, how fast is it going?

 

 

it seems to me you should find that easy since

you know the earth is going 30 km/second

and that the acceleration falls off as square of distance

 

like

 

[math]\frac{V_J^2}{R_J} = \frac{1}{25}\frac{V_E^2}{R_E}[/math]

 

or you may have a still neater way to deal with it, but in any case

it should be easy to say how fast Jupiter is going simply from the fact that it is 5 times farther out. I hope so anyway

 

BTW to prevent any possible confusion (I know you realize this J5 but in case other people drop in) I should say that the R here is the orbit radius not the radius of the planet!

[Johnny5]

I like that answer' date=' personally. I think it is right. Anybody want to double check what J5 says?

 

I think that the acceleration towards the sun, due to the sun's gravity, at our present distance away from it, is 0.006 meters per second square like what he says. Would be nice if someone else would go thru the arithmetic though, to catch any careless error.[/quote']

 

I wasn't positive it was right, in fact to be honest, today I no longer remember the question. I think I was trying to figure out, what the centripetal acceleration of the earth is, around a fixed sun.

 

Any analysis requires that you perform it in the right reference frame, in order to make sense, i know that for sure.

 

Correct me if I am wrong, but in order to use v^2/R, the R has to be the distance from the center of a circular orbit. Since the sun is so massive, the center of mass of the earth-sun system is 'probably' inside the sun. But see I really don't know this. If I just assume that the center of mass of the earth-sun system is the center of the sun, and use the numbers you gave, and just chug without thinking, I get that answer.

 

But doing physics this way will only work if you are carrying out the analysis in the right reference frame. That's the most important thing of all.

[Johnny5]

1 second later he will be travelling at 9.78 m/s (in the downwards direction) but will have dropped only 4.89 metres. His horizontal/rotational velocity/vector should exceed that of the Earth.

 

Now that I think about it, I answered that whole part wrong. All I wanted to figure out was how much the spot underneath him moves to the right, in the time it takes him to strike that ground. Thats what I wanted to calculate.

 

But in hindsight, the acceleration of .033 m/s^2 was centripetal accelaration, not rightwards acceleration. The earth is not accelerating to the right in his frame. In fact, the rotational velocity of the earth in the appropriate frame is V, where centripetal acceleration = v^2/r

[Johnny5]

But if the person is stationary with respect to the ground before the freefall, that means he is moving with the same angular speed.

 

Yes I know that, how silly of me to ignore it. The centripetal acceleration isn't rightwards, I did the problem wrong.

 

Let me try it again.

 

Here is the problem fresh.

 

On the real earth...

 

There is a pole stuck in the ground, at the equator.

 

The acceleration of earth's gravity at the equator is 9.78 m/s^2 by direct measurement.

 

Let the height of the pole be 9.78 meters.

 

There is a man holding onto the pole at the top of it, so that he is 9.78 meters above the ground below him. In a moment he is going to let go of the pole, and free-fall to the earth.

 

Ignore air, assume he falls in a vacuum. We built a big tube, evacuated the air, and he and the pole are inside. But still the experiment is taking place on the real earth.

 

Now, there are two reference frames. One which is attached to the man, and one which is attached to the base of the pole. Let the man have set up a coordinate system in his rest frame, with his center of inertia at the origin, and let his arms be extended straight out so that his whole body forms a letter t. Let his right arm point in the i^ direction let his left arm point in the -i^ direction, let his head point in the j^ direction, let his feet point in the -j^ direction, and let his eyes which are looking straight at the ground be looking in the k^ direction. This coordinate system travels with the man as he falls. Let the pole frame have its origin located at the base of the pole, let its x and y axes be parallel to the man's x,y axes, and let its +z axis point at the man. So at least initially, the z axes of both frames coincide.

 

If these reference frames are coupled (z axes of both frames coincide while he falls), then the man will hit the spot where the pole is stuck in the ground. On the other hand, if at the moment he lets go these frames decouple, then in the rest frame of the man, the spot where the pole is will actually move some total horizontal distance D, in the time it takes him to free-fall, in addition to the vertical distance it moves in his rest frame (which is 9.78 meters).

 

Note: Relativistic length contraction effects certainly do not apply to the pole before the man lets go, because the pole never moves in its own rest frame, so in that frame it always has that length, and initially the man is at rest in the pole rest frame, so that he agrees that if he lets go of the pole, then he will fall 9.78 meters. Those who want to relativistically analyze this problem feel free.

 

And equivalently, if at the moment he lets go, the frames decouple, then in the rest frame of the pole, the man will actually move slightly to one side.

 

Also, there is of course motion in the falling direction. In the rest frame of the man, the surface of the earth will appear to accelerate towards him at 9.78 m/s^2. In the rest frame of the pole, the man will accelerate towards the ground at 9.78 m/s^2... the up/down acceleration goes without saying.

 

If in either frame, the man does not move straight to the ground, then the two frames I am switching back and forth between, decoupled the moment he let go, and the path of the other thing in either frame, is almost diagonal, but actually I think Galileo would have said slightly parabolic.

 

But the whole issue is whether or not he hits the spot directly underneath him.

 

First let me approximate the time of free fall. I know that either he is going to move straight down in his rest frame, or the pole will move slightly to one side as he falls.

 

By experience, we all know that if there is any slight motion to one side, that it is imperceptible. That doesn't mean its not there, it means exactly what it means. All you have to do is jump out of your tree house, to know.

 

So as a first approximation to the total time of free fall, let me assume he falls in a straight line right down to the ground.

 

So I am carrying out the analysis in the rest frame of the pole.

The man is presumed to be linearly accelerating at 9.78 m/s^2 in this frame.

 

So we may use the kinematical equations for constant acceleration.

 

By definition:

 

[math] a \equiv \frac{dv}{dt} [/math]

 

Multiply both sides by dt (dt, the time differential, is measured in the pole's rest frame ) to obtain:

 

[math] a dt = dv [/math]

 

Let t1 be the moment in time at which he lets go of the pole, and let t2 be the moment in time the ground collides with him. Let v denote the relative speed between the ground and him at all moments in time. Before he lets go of the pole, and including the moment he lets go, the relative speed v=0. After he lets go v begins to increase. So let v1 denote the relative speed of the man to the ground at moment in time t1, initially the man isn't moving relative to the ground, so that v1=0 in computation. Let v2 denote the speed that the ground impacts the man, which as was said occurs at moment in time t2. Therefore:

 

[math] \int_{t=t1}^{t=t2} a dt = \int_{v=v1}^{v=v2} dv [/math]

 

And initially, the relative speed of the man to the ground is zero, so that:

 

[math] \int_{t=t1}^{t=t2} a dt = \int_{v=0}^{v=v2} dv [/math]

 

Now, we are treating the acceleration during free fall as being constant. This isn't true, and we can prove that it's not true using the Newtonian gravity formula, but we are assuming it is true, in order not to make the calculations too complex. Remember, I am just trying to approximate the time of free fall.

 

Since the acceleration is assumed constant during the man's fall, we can pull it out of the integral to obtain:

 

[math] a\int_{t=t1}^{t=t2} dt = \int_{v=0}^{v=v2} dv = v_2 - 0 = v_2[/math]

 

And

 

[math] a\int_{t=t1}^{t=t2} dt = a(t_2 - t_1) [/math]

 

t2-t1 is the amount that "time has changed" according to a clock at rest in the pole rest frame. Assume that it is measured in seconds, and denote it by [math] \Delta t [/math]. Therefore we have:

 

[math] a \Delta t= v_2[/math]

 

And we said that at the equator, a was measured to be 9.78 meters per second squared, therefore:

 

[math] 9.78 \Delta t= v_2[/math]

 

Now, we can write what happens as a moment to moment true statement, if we stipulate that at the moment the man lets go of the pole, the rest clock reads zero seconds. Let the relative speed at any moment in time after he lets go of the pole be denoted by v. Then we have:

 

[math] a\int_{t=t1}^{t=t} dt = \int_{v=0}^{v=v} dv = v - 0 = v[/math]

 

From which it follows that:

 

[math] a(t - t_1) = v - v_1 [/math]

 

And v₁ = 0, and at the moment he lets go t₁=0 therefore:

 

[math] at= v [/math]

 

The previous statement is true in the rest frame of the pole, at all moments in time after the man releases the pole. Since the acceleration of gravity at the equator a, is 9.78 meters per second squared, let the time be measured in seconds, so that the instantaneous speed will have units of meters per second. Therefore, in the rest frame of the pole we have:

 

[math] 9.78t= v [/math]

 

Now, his instantaneous speed (which was defined by Galileo I believe) is given by:

 

[math] v \equiv \frac{dh}{dt} [/math]

 

Therefore:

 

[math] 9.78t= \frac{dh}{dt} [/math]

 

Multiplying both sides of the previous statement by dt we have:

 

[math] 9.78t dt = dh [/math]

 

Now, we integrate from the moment the man lets go of the pole and starts falling towards the earth, to the moment the ground collides with him. His height is 9.78 meters above the earth when the clock reads t=0. Therefore we have:

 

[math] \int_{t=0}^{t=t} 9.78t dt = \int_{h=9.78}^{h=0} dh [/math]

 

From which it follows that:

 

[math] 9.78 \int_{t=0}^{t=t} t dt = 0 - 9.78 = -9.78 [/math]

 

Using the integral calculus we can show that:

 

[math] \int_{t=0}^{t=t} t dt = \frac{t^2}{2} [/math]

 

From which it follows that:

 

[math] 9.78 \frac{t^2}{2} = 0 - 9.78 = -9.78 [/math]

 

The quantity on the LHS must be positive since t² must be positive, and the quantity on the RHS is negative, therefore we have introduced a minus sign error. The only place it can be located is where we introduced dh, for a change in height.

 

Since the object is falling down, the height is decreasing, not increasing, therefore we should have written:

 

[math] 9.78t= -\frac{dh}{dt} [/math]

 

In which case we would have obtained the following true statement in the rest frame of the pole:

 

[math] 9.78 \frac{t^2}{2} = 9.78 [/math]

 

Dividing both sides by 9.78 we obtain:

 

[math] \frac{t^2}{2} = 1 [/math]

 

Now multiplying both sides by 2 we obtain:

 

[math] t^2 = 2 [/math]

 

Now taking the square root of both sides of the statement above we obtain:

 

[math] t = \sqrt{2} [/math]

 

The square root of two is greater than the square root of one, and less than the square root of four. The square root of 1 is equal to 1, and the square root of four is equal to two. Let us guess that the square root of two is equal to 1.5, (1.5)(1.5) = 2.25, therefore we have guessed too high. Let us guess 1.4, (1.4)(1.4)= 1.96, so 1.4 is accurate to 1 decimal place.

 

The conclusion is, that if the man falls straight down, rather than a parabolic path, that the total time of free fall is 1.4 seconds as measured by a clock in the rest frame of the pole, which is located at the equator of the earth.

 

We have introduced another error, which is that we treated time as a continuous variable, rather than a discrete variable. That means that between any two moments in time there are an infinite number of moments in time, and that is impossible.

 

We also assumed that the man falls in a straight line and he does not, the exact trajectory which the center of inertia of the man traces out in the rest frame of the pole remains to be determined. Is it a parabola?

 

Additionally, a real man is a large body, and even as he distorts his arms and legs, his center of inertia remains inside of him. So his center of inertia will never actually reach the ground. The treatment here is more accurate for a small particle, whose radius is small in comparison to the height it is dropped.

 

We perform the experiment N times, and compute the average to be 1.4 seconds, so our mathematical treatment of the problem thus far isn't too horrendous, even though we have used numerous false statements to arrive at the value of "square root of two" seconds.

[Johnny5]

BTW J5' date=' jupiter like the earth has a roughly circular orbit and is 5 times farther from the sun, how fast is it going?

 

 

it seems to me you should find that easy since

you know the earth is going 30 km/second

and that the acceleration falls off as square of distance

 

like

 

[math']\frac{V_J^2}{R_J} = \frac{1}{25}\frac{V_E^2}{R_E}[/math]

 

or you may have a still neater way to deal with it, but in any case

it should be easy to say how fast Jupiter is going simply from the fact that it is 5 times farther out. I hope so anyway

 

BTW to prevent any possible confusion (I know you realize this J5 but in case other people drop in) I should say that the R here is the orbit radius not the radius of the planet!

 

Your question is, how fast is jupter going.

 

Well that question only makes sense if you specify what reference frame you are defining the speed in.

 

In order for this question to make sense, you want it to be the same frame you asked me that question about earth in, because that is the frame in which earth has the 30km/s speed you keep referring to.

 

Now, in a reference frame attached to the earth, the earth has no speed, so it's not the earth frame we are talking about here.

 

And it's not really the sun's rest frame either, what we need is the CM frame of our solar system really, but I'm not sure where that is in relation to the center of mass of the sun.

 

For the sake of simplicity, assume that the CM of our solar system is the CM of the sun.

 

Now, imagine a vector that always points from the CM of the sun to the CM of the earth. Let this vector define an axis of Sun-frame.

 

So in Sun-frame the earth isn't moving at all; so Sun-frame cannot be the frame we calculated the speed of earth in.

 

We need to choose a frame which is at rest relative to the fixed stars i think. I hear everyone talking about it, but it's not clear how we would make measurements designed to prove we are at rest in such a frame.

 

Let us say that our sun is perfectly spherical. That it isn't buldging in any cross section. Let us infer from this non-bulding (which isn't true since the real sun buldges) that any reference frame in which a point on the surface of the sun is at rest, is an inertial reference frame. Someone please correct that if it's wrong.

 

So now, watch the motion of the earth from such a frame. In this frame, the earth is orbiting the sun, with some speed V, and we previously calculated its centripetal acceleration.

 

The new question now, is what is the speed of jupiter in this same frame.

 

I will assume that jupiter and earth orbit the sun in the same plane, and that both orbits are perfectly circular in this frame (all false I know).

 

At any rate, let it be the case that the orbital radius of Jupiter is 5R_e, where R_e is the orbital radius of earth.

 

We are asked to compute the speed of the planet jupiter in this frame. In other words, what is the speed of the center of inertia of Jupiter in this frame.

 

You say that acceleration falls off as the square of distance, but then give a formula where it falls off with the distance. I do not understand why you did that.

 

I will just try to answer the question correctly.

 

I am going to use Newton's formula for gravity.

 

[math] \frac{GMm}{R^2} = m \frac{v^2}{R} [/math]

 

Where M is the mass of the sun, G is the gravitational constant, and m is the mass of a planet in circular orbit around the sun.

 

The RHS is the formula for centripetal force.

 

In the case of Jupiter, we want to use the mass of Jupiter which is far greater than the mass of earth.

 

I had to go back and read an old post of Martin's

 

it is a handy thing to remember that the earth is always going about one tenthousandth of the speed of light in its roughly circular orbit around sun

 

 

(it gives a convenient handle on the doppler effect of the earth's own orbit speed)

 

 

what is the acceleration due to the sun's gravity at this distance where we find ourselves

 

 

===========

hmmmm looks like I have to go out for the morning

 

well the earth is going 30 km/sec and it is some 150 million km from sun

so it should be possible to find our sunwards acceleration that is keeping us in a roughly circular orbit. Is this accelerations greater or less than the 9.8 earthwards acceleration we all feel standing here in our shoes?

 

In that post you gave the distance from the center of inertia of the sun' date=' to the center of inertia of earth as 150 million kilometers. You then stated that Jupiter is five times further away from the center of inertia of the sun than earth is. Therefore, the center to center Sun-Jupiter distance is:

 

[math'] 5 (150 x 10^9 meters) = 750 billion meters [/math]

 

If I now know the sun's inertial mass, I can solve the problem, since the inertial mass of Jupiter will cancel out. In other words:

 

[math] \frac{GM_{sun}}{R} = v^2 [/math]

 

I do not currently know the inertial mass of the Sun, but suppose that the sunward centripetal acceleration of the earth calculation I made earlier is correct. That computation was .006 m/s^2, toward the sun, in the frame I am now trying to compute Jupiter's speed in. So in this frame the following statement is supposedly true:

 

[math] \frac{GM_{sun}m_{e}}{R^2} = m_e \frac{v^2}{R} = m_e (.006)[/math]

 

From which it follows that:

 

[math] \frac{GM_{sun}}{R^2} = (.006) [/math]

 

And the R in the formula above, is the center to center distance between sun and earth, which was given as 150 million kilometers. So that we have:

 

 

[math] GM_{sun}= (.006) (150 million km)^2 [/math]

 

Where the LHS is the gravitational constant, times the inertial mass of the sun. For Jupiter we have:

 

[math] \frac{GM_{sun}}{750 billion meters} = v^2 [/math]

 

where v is the orbital speed of Jupiter, which we were asked to find. Therefore:

 

[math] GM_{sun} = 5(150 million km) v^2 [/math]

 

Therefore:

 

[math] (.006) (150 million km)^2 = 5(150 million km) v^2 [/math]

 

Therefore:

 

[math] (.006) (150 million km) = 5 v^2 [/math]

 

Therefore:

 

[math] (.006 m/s^2) (30 X 10^9 m) = v^2 [/math]

 

Therefore:

 

[math] v^2 = 1.8 X 10^8 m^2/s^2 [/math]

 

From which it follows that

 

V = 13,416 meters per second, or 13.4 km/s

 

I could have made a computational error easily though.

[Martin]

From which it follows that

 

V = 13' date='416 meters per second, or 13.4 km/s

 

I could have made a computational error easily though.[/quote']

 

 

that looks right!

 

I will explain my own reasoning in a moment (just as you have explained yours, which I am glad for) but first let me check the answer.

 

basically since jupiter is 5 times farther away it should be slower by a factor of sqrt(5)

 

so we want to divide 30 km/sec by the sqrt(5)

 

whoah! 13.416 km/sec just as you say!

 

now we can make all the corrections we want, it is not EXACTLY 5 times farther, it is 5.2 times farther, so we can make that little correction

 

and also the earth mass really is pretty much negligible so it is OK to approximate by neglecting it but Jupiter mass is maybe not quite so negligible so if we made all the corrections blah blah blah.

 

but I think it is always best to go for the approximate answer which is around 13 km/s or 13.4, and then do it more precisely later if you happen to want to. which in this case I dont

 

very glad you got 13.4, Jupiter will be happy when we tell him, it will reassure him to know he is going the right speed