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What gives you the idea that it would?

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What gives you the idea that it would?

 

because of the gravity by the sun

 

surely the gravity by the sun is strong enough to eventually draw in Mercury and Venus is it not ?

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As long as a planet has good velocity, it will maintain its distance from the sun. If you could slow down Mercury in its orbit, it would spiral into the sun. However, how are you going to do that? Friction with particles in space will do it to a very minor extent, but this is a whole planet we are talking about, with utterly immense momentum. A little friction will not be enough to do it.

 

We can expect Mercury to continue in something close to its current orbit until the sun goes red giant in about 5 billion years.

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surely the gravity by the sun is strong enough to eventually draw in Mercury and Venus is it not ?

Surely not.

 

The place to start is with the two body problem in Newtonian physics. If the Sun and Mercury were the only bodies in the solar system, Newton's law of gravity was perfectly correct, the Sun's mass was constant, and the Sun was perfectly spherical, then Mercury would orbit the Sun in a never-changing ellipse. While gravity does indeed make Mercury constantly accelerate toward the Sun, it's motion normal to its radial vector keeps it from falling into the Sun.

 

None of those "ifs" is true. The other planets, particular Jupiter, act to perturb Mercury's orbit. These perturbations change the orientation and shape of the orbit (and do so by quite a bit over astronomical time). However, they do not change the orbital energy by much at all and the changes are cyclic.

 

Newton's law of gravity is only approximately correct. General relativity does a better job, and it predicts a slight change in the orientation of Mercury's orbit. This slight change, the relativistic precession, explained a discrepancy that astronomers could not explain before GR. This is one of the key reasons general relativity was accepted so quickly in the scientific community. This precessional advance is just a change in the orientation of Mercury's orbit. It will not make Mercury fall into the Sun. General relativity also predicts that Mercury will emit gravitational waves (akin to electromagnetism), and this will make Mercury's orbit decay. The decay is extremely small, however. Mercury is moving far too slow for gravitational radiation to have any sizable impact on its orbit over the life time of the solar system. The decay from gravitational radiation is immeasurably small.

 

The Sun is not perfectly spherical. The deviation from sphericalness is very small, and once again only acts to change the orientation and shape of Mercury's orbit. This affect is even smaller than the relativistic precession.

 

Finally, the Sun does not have a constant mass. It loses mass in the form of electromagnetic radiation and solar wind. This mass reduction means that Mercury will slowly move away from the Sun. This outward drift resulting from solar mass loss is in fact larger than the decay resulting from gravitational radiation. Mercury is moving away from the Sun (at an extremely small rate).

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why doesn't Mercury simply fall into the sun ? eventually ?

 

Basically for the same reason none of the other planets fall into the sun, their orbital velocity.

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Surely not.

...

Mercury is moving away from the Sun (at an extremely small rate).

 

but is Mercury actually doing so

 

is there physical dynamic evidence for what you suggest ?

Edited by Sayonara³
quoted whole post for no real reason

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but is Mercury actually doing so

 

is there physical dynamic evidence for what you suggest ?

I should have said "theoretically".

 

We directly observe sunlight and solar wind, thereby providing indirect evidence that the Sun is losing mass. The combined effect is a decrease of 9×10-14 solar masses per year. To conserve angular momentum, Mercury's orbit (specifically, its semi-latus rectum) must increase by a factor of 9×10-14 per year, or a paltry 5 millimeters per year.

 

This is far too small to be observable.

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The reason Mercury does not not fall into the sun is because of its centripital acceleration. Consider an object moving in a circle of radius r with constant angular velocity. The tangential speed is constant, but the direction of the tangential velocity vector changes as the object rotates.

 

Here's a good example, if your in a car when it turns a corner the top half of your body wants to move away from the centre of rotation, tilt your body toward the turn until you reach equilibrium. This is the same as the orbital status of Mercury except in this case Mercury's velocity acts like your angle of tilt, not to high and not to low.

 

The direction of the centripital acceleration is always inwards along the radius vector of the circular motion.

 

The magnitude of the centripetal acceleration is related to the tangential speed and angular velocity.

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The reason Mercury does not not fall into the sun is because of its centripital acceleration.

That's not quite right. Here is a thought experiment that shows a counterexample. Put a pulley wheel on a pivot so the pulley wheel can rotate around the pivot. Attach a string to a mass and string the string through the pulley. Now swing the string so the mass undergoes circular motion. Now start pulling the string through the pulley. The rock will hit the pulley when the string gets short enough. Note that at all times the force on the rock is a central force.

 

The reason Mercury does not fall into the Sun is because (a) Mercury has a non-zero angular momentum with respect to the Sun, (b) ignoring relativistic effects, gravity is an inverse square central force, and © perturbations such as those due to relativity and other planets are small.

 

 

The tangential speed is constant,

That is only true for a circular orbit. Mercury does not have a circular orbit; it's not even close (eccentricity=0.206).

 

The magnitude of the centripetal acceleration is related to the tangential speed and angular velocity.

No. The magnitude of the acceleration is related to the distance only. That is how gravity works.

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Think of it this way, if you have a ball tied to a string then if the ball is not moving then the centripital force is zero. When you start moving it around then it starts gaining centripetal acceleration which is why it starts moving away from the ground (spinning horizontally)

The equation, F = mv^2 / r , is for centripetal force. You can rearrange this to, F = m.(2pi.r/t) / r, where t is time interval for one cycle and r is the distance. Three factors, mass, velocity and radius( distance from centre of orbit)

 

It is the centripital force that stops it from plunging back into the sun. That's how satellites orbit the Earth and dont fall back into the atmosphere

 

The centripital acceleration of Mercury is equal to the gravitational pull of the sun

 

This is the centripital force of Mercury

F = ( mv^2 ) / r = ( 3.3e23 x 47930.4435^2 ) / 5.7872e10 = 1.3071e22 N

 

 

This is the Sun's Gravitational Field Strength at Mercury

g(sun) = (G.M) / r^2 = (6.67e-11 x 1.98892e30) / (5.7872e10)^2 = 0.03961012072 N/kg

 

 

This is the weight of Mercury relative to the Sun's Gravity

W = m.g = 3.3e23 x 0.03961012072 = 1.3071e22 N

 

Centripital force - weight of Mercury = 0.0000 N to 4 d.p

 

It is an elliptical but these rules still apply for when a planet orbits a star.

 

I had a year 12 project on the same thing, about centripetal force and how satellites are put into a stable orbit

 

 

http://theory.uwinnipeg.ca/physics/circ/node7.html

Edited by einsteinium

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No, it is the centripital force that stops it from plunging back into the sun. That's how satellites orbit the Earth dont fall back into the atmosphere

 

The centripital acceleration of Mercury is equal to the gravitational pull of the sun

You are not understanding what I wrote. You claimed that it is the centripetal force that stops Mercury from falling into the Sun. That statement taken by itself is incorrect, for two reasons.

 

First, any centripetal acceleration (a kinematic rather than dynamics concept) will not do. The centripetal acceleration has to take the correct form. Orbits are not stable if, for example, the centripetal acceleration is of the form constant+f®. Things get even more restrictive when moving from kinematics (centripetal accelerations) to dynamics (central forces). While circular orbits are metastable for any central forces of the form krn, they are stable only if n>-3. Moreover, closed orbits only appear if f=kr (e.g. Hooke's law) or if f=k/r2 (e.g. gravitation).

 

Secondly, the object in question needs a non-zero angular momentum with respect to the central body. An object with zero tangential velocity relative to the Sun will fall into the Sun.

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The centripetal force cancels out the force due to the Sun's gravity making it relatively weightless.

 

It is not a perfect orbit as I have recently stated. The velocity that Mercury moves at wants to throw it out of orbit but the Sun's gravity keeps pulling it back in.

 

For every action, there is an equal but opposite reaction. This is what I'm trying to say

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The centripetal force cancels out the force due to the Sun's gravity making it relatively weightless.

"Centripetal" means "toward the center". Another name for centripetal force is central force. Gravitation is a central force, and it certainly does not cancel itself. I suspect you might mean centrifugal force, and that of course is not a real force. It is instead a fictional force.

 

For all practical purposes, the only real force (Newtonian mechanics) acting on Mercury is that of gravitation. Other forces include drag against the interplanetary medium and the solar wind and solar radiation pressure; these are tiny. Those other forces don't come close to canceling that of gravity.

 

The term "weight" has several conflicting meanings. Legally, weight is a synonym for mass. To a lawyer and to a merchant, the contents of a can of vegetables weighs exactly the same on the Earth, on the Moon, and on the International Space Station. Forgetting the legal definition (who cares?), the term "weight" has multiple meanings even within the context of physics.

 

One definition of weight is simply mass times the acceleration due to gravity. In this sense, Mercury is anything but weightless. Then again, an astronaut in the space station is anything but weightless by this definition. Their weight is about 90% that of what they weigh on the surface of the Earth.

 

So why do we say astronauts are "weightless" when they are in orbit? Simple: Their weight as measured by a spring scale would be near zero. A spring scale doesn't sense gravity. This leads to the second definition of weight: the net force acting on an object except for gravitational forces. For a person standing on the surface of the Earth, these two definitions of force are nearly identical in magnitude and nearly opposite in direction. For a person in the space station, the two definitions are quite different.

 

A third definition of weight is mass times the acceleration measured by a perfect accelerometer. This is essentially the same as weight as measured by a spring scale. The reason a scale or an accelerometer do not and cannot sense gravity is deep. Scales cannot sense fictitious forces such as centrifugal force. Gravity itself is a fictitious force in general relativity. In Newtonian mechanics, gravitation is a real force. You will do fine looking at gravity as a real force until you start studying general relativity. If you are a 12th year student, that won't happen for three of four years. I'll stick with the Newtonian interpretation for the rest of this post.

 

It is not a perfect orbit as I have recently stated.

What do you mean by this? In particular, what is a "perfect orbit"?

 

For every action, there is an equal but opposite reaction. This is what I'm trying to say
You appear to have a misconception of Newton's third law. Not surprising; most pre-university physics instructors themselves have a poor understanding of Newton's third law and they impart this misunderstanding on their students.

 

An example of a misapplication of Newton's third law: A person standing on a spring scale. Many instructors teach that the downward gravitational force by the Earth on the person and the upward force by the scale on the person is an example of Newton's third law. This is completely wrong. A third law pair always involves the same force acting on two different bodies. The cited example is wrong because it involves two different forces acting on one body. Moreover, the forces are not equal but opposite. The person is undergoing uniform circular motion at the rate of one revolution per day.

 

There are several third law force pairs involved when considering a person standing on a scale. The Earth exerts a gravitational force on the person. The third law counterpart: The person exerts an equal-but-opposite gravitational force on the Earth.

 

So what about the scale? That is the normal force (electrostatic repulsion). The scale is exerting an upward force on the person. This upward force keeps the person from sinking into the scale. The third law counterpart: The person exerts an equal-but-opposite normal force on the scale. This in turn compresses the scale's spring. The reading displayed on the scale is this compression, translated from distance to force by means of Hooke's law.

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I know what your saying and I agree with you.

 

When I stated "not a perfect orbit" I'm saying that it's an elliptical orbit and the velocity of Mercury during its orbit and its orbital radius will vary which contributes to the impairments of the forces acting upon Mercury at any time.

 

I know all about weight, whenever I hear someone say "what is your weight?" I feel like laughing when they say it in kg somehow. Whenever someone ask me this I answer them with my weight in Newtons or Dynes.

 

I already know about the "toward centre acceleration" which is what I'm saying. An astronaut is RELATIVELY weightless meaning that the gravitational field strength force is equal to his centripital acceleration, his normal force is zero. This doesnt mean that he is weightless, it just means that the gravitational field strength is equal to his centripital acceleration which means they theoretically give him an apparent weightlessness.

 

Apparent weightlessness and true weightlessness

Any person or object will experience APPARENT WEIGHTLESSNESS when falling with an acceleration equal to the gravitational field strength.

 

TRUE WEIGHTLESSNESS occurs only in deep space, out of reach of the gravitational field of anything with a mass or when orbital radius tends to infinity, g = G.m / r^2

Edited by einsteinium

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II know all about weight, whenever I hear someone say "what is your weight?" I feel like laughing when they say it in kg.

Those people are correct. Your argument that weight is a force carries less weight than theirs that it is mass. Legally and colloquially, weight is mass. Our scientific definitions of weight are jargon. The people who write dictionaries always view jargon as having less weight than the common use of a word.

 

I already know about the "toward centre force" which is what I'm saying. An astronaut is RELATIVELY weightless meaning that his weight force is equal to his normal force but times negative one.

The gravitational force and normal force acting on an astronaut nearly cancel when the astronaut is standing on the surface of the Earth. When the astronaut is in orbit, the normal force is essentially zero. So are you saying an astronaut is weightless when he is on the ground, but not when he is on orbit? If not, you need to rethink what you just said.

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Re weight of astronaut.

 

The 'weightlessness' is an illusion based on the fact that the astronaut is falling.

 

Imagine an astronaut in orbit 1000 kms above the Earth. He appears to be weightless, but this is only because he is moving. He is falling at the same rate as the apparatus used to measure his weight, which means the apparatus will give an incorrect reading of zero.

 

Now imagine the same astronaut 1000 km above the Earth - but this time he is standing on something like a massively taller hypothetical version of the Eiffel Tower. The astronaut is not falling because he is standing on a firm surface. In this case, the astronaut has weight, and the apparatus will measure that weight as a high percentage of his weight at Earth sea level.

 

What DH has been saying is correct, but sometimes unnecessarily difficult to understand. A circular orbit is understood simply by accepting that the object in orbit is falling, but the rate of fall is balanced by the fact that it moves out and away due to its orbital motion. In one minute, it may fall a kilometre, but in that minute it moves tangentially to the object it orbits, and thus moves out by one kilometre. The fall and the movement out balance.

 

For an elliptical orbit, the distance fallen and distance moved out will not be the same, minute by minute, but average out equal over one whole orbit.

 

Orbits change. If an object is made to slow, like low orbit Earth satellites that contact the outer atmosphere and are slowed by friction, it will spiral inwards to the object it orbits. If it is made to speed up, it will spiral out further from the object it orbits.

 

A good example of the latter is Earth's moon. Earth's rotational energy is being slowly transferred to the moon by tidal interaction. The Earth's rotation is slowing, and the moon, in response, moves faster around the Earth. Over a period of millions of years, the moon moves further out from the Earth. The moon's orbit takes longer since it has to travel much further. Both the slowing of the Earth's rotation and the moon's longer orbit time can be, and have been, measured using highly accurate atomic clocks.

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Re weight of astronaut.

 

The 'weightlessness' is an illusion based on the fact that the astronaut is falling.

It's not an illusion. It is very real. You are assuming weight is mass times gravitational acceleration. This is not the only definition of weight used by physicists (see my prior posts), and IMHO it is not a very useful one. Why not? Simple. No device can be constructed to measure weight defined this way. Weight in general relativity is mass times the reading from an ideal accelerometer. This is a useful definition because it suggests a measurement scheme. With this definition, an astronaut in orbit is weightless.

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To DH

Weight as opposed to mass is the effect of gravity on an object. If gravity is less, weight is less, while mass will be constant.

 

An astronaut in orbit appears to be weightless, but gravity is nevertheless acting on his body. The result is that the astronaut is falling. It is essentially no different (assuming no effect from air friction) to a person in free fall before he opens his parachute. In both cases they have weight since gravity is exerting an infuence (acceleration). In both cases, a devise for measuring weight will show zero weight, since the devise is also falling, and at the same rate.

 

In both cases, if you were to exert a deceleration to balance the acceleration under gravity you would see the weight suddenly appear to return, and become measurable with simple scales.

 

A good example is a base jumper. Standing on the jump site, he has weight. After jumping, and in free fall, he appears to be weightless, but the effect of gravity has not changed in quantity - just a qualitative effect, now exerting its effect as acceleration.

 

If an astronaut is sufficiently far from the Earth, the effect of gravity is small, and he is said to be in a micro-gravity environment. Not a zero gravity environment.

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To DH

Weight as opposed to mass is the effect of gravity on an object. If gravity is less, weight is less, while mass will be constant.

Read the previous posts, Lance. That is not the definition of weight used in advanced physics classes (weight is mass times accelerometer reading) and even in some intermediate physics classes (weight is what a spring scale measures).

 

An astronaut in orbit appears to be weightless, but gravity is nevertheless acting on his body.

Gravity isn't a real force. It's a pseudo force.

 

Pseudo forces (centrifugal force, coriolis force, inertial force) have two things in common. One is that the magnitude of such a force acting on an object is proportional to the mass of the object. Gravity: The gravitational force acting on object is proportional to the mass of the object.

 

The second thing pseudo forces have in common is that the pseudo force acting on some object cannot be measured by any sensor on that object. Pseudo forces are not intrinsic forces. They instead are a consequence of the observer. Gravity: There is no way to measure the gravitational force acting on an object. This is a direct consequence of the equivalence principle.

 

It is essentially no different (assuming no effect from air friction) to a person in free fall before he opens his parachute. In both cases they have weight since gravity is exerting an infuence (acceleration).

Correct.

 

 

In both cases, a devise for measuring weight will show zero weight, since the devise is also falling, and at the same rate.

Also correct.

 

In both cases, if you were to exert a deceleration to balance the acceleration under gravity you would see the weight suddenly appear to return, and become measurable with simple scales.

Also correct. However, that deceleration results from some real force acting on the object. The scale is measuring those other forces. It is not measuring the gravitational force. The gravitational force is unmeasurable. From a Newtonian perspective, the reason is that the gravitational force cannot be shielded. From a general relativistic perspective, the reason is even simpler: Gravity is not a real force.

 

The rest of your post indicates your thinking is too Newtonian. While gravity is a real force in Newtonian mechanics, it is not a real force in general relativity. Newtonian mechanics is not the be-all and end-all to physics. In physics, the model that more accurately represents reality is the winner. General relativity does a better job of explaining reality than does Newtonian mechanics. Gravity is not a real force.

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To DH

Everything you say in your post is, of course, correct. However, the point I am making is that an astronaut in orbit still has weight, since there is gravity acting on him. In fact, it creates an acceleration which is measurable, and the concept of mass times accelerometer reading applies.

 

When falling through a substantial gravity field, the concept of weightlessness is invalid.

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An astronaut in orbit still has weight if you use the tautological definition of weight being mass times gravitational force. How many times do I have to say this: There are multiple, conflicting meanings of "weight". You are stuck on just one.

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