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What causes orbit speed

  1. Dec 4, 2009 #1


    At radius 1: Acceleration Due to Gravity (ADG) is 16 times as strong as at radius 4.
    The object at radius 1 has 4 times so much KE and 4 times so little time to change the course (into a continues circular orbit) hence ADG must be 4*4 times so strong, = ADG 16 at radius 1,- to keep the object in circular motion. - So far so good.....

    BUT the point is:
    that the speed of the object at radius 1 is only double (*2) - even though that gravity ADG is *16 – WHY.

    It seems that there is a high price to pay for keeping the object in orbit, and that price is lower orbit speed. – Right ? - Or what's going on here ?

    Additional question

    According to GR gravity is caused to curvature of space, not to a force..
    Or an object is not following an orbit because of a force or because of energy, but because of a property of space.

    According to Newtonian gravity , Gravity is a force. = F
    How is this possible? Are we ridding both a bull and a horse at the same time?

    1. Is gravity a force, - or is gravity not a force?
    2. And if gravity is a force, how can a force pop up from space ? - I mean without energy ?
  2. jcsd
  3. Dec 4, 2009 #2


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    Orbit speed (in a circular orbit) around A massive object (so you don't have to consider co-orbiting) can be easily derived from Newton's laws and a little knowledge of circular orbits.

    In order to maintain a circular orbit, the acceleration must be:

    From Newton's law of gravitation, and F=ma:

    [tex] F=\frac{GMm}{r^2}=ma=m\frac{v^2}{r}[/tex]

    Now solving for v:


    So you can see if r is quadrupled, the speed only goes down by a factor of 2 (because of the square root).
  4. Dec 4, 2009 #3

    I understand the math…
    But can you say: is that there is a high price to pay (orbit speed) for keeping the object into a circular orbit.
    I mean the force that maintains the orbit speed is in the end: Acc.. due to Gravity (ADG)
    ADG is increasing much more than the orbit speed increases...
    Is there a lost of speed for continuously to “bend” the velocity direction to a circular orbit…?
  5. Dec 4, 2009 #4


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    I don't think I quite get what you are asking...

    The acceleration is determined by Gravity, the velocity is determined as the velocity needed to keep the circular orbit as I have derived earlier. I don't know what else to say...
  6. Dec 4, 2009 #5


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    The short answer is conservation of angular momentum, as already noted [with the math]. Orbital velocity is merely the speed a planet, or other massive object, needs to acheive to avoid falling into the sun. Planets are the residue of mass that sucessfully achieved this velocity.
  7. Dec 4, 2009 #6

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    This picture is incomplete. There are elliptical orbits, too. Suppose in that picture you brought an object out to the fourth circle and made the object's velocity zero with respect to the planet. The resultant path is what some jokingly call an orthogonal orbit. The object will fall straight toward the planet.

    So how does an object get into a circular orbit? The answer is: Something must give it the precise velocity needed to be in a circular orbit. Once it has that precise velocity, it will continue orbiting forever. But it has to get that initial momentum from somewhere.
  8. Dec 4, 2009 #7


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    Your first question has been answered, I'll have a stab at the second.

    The Newtonian and Einsteinian viewpoints are different, but each is internally consistent. Take a very simple situation; a cup of coffee sitting on your desk. There are two ways of looking at this.


    The force due to gravity between the cup and the centre of the Earth applies a force downwards on the cup. In the abscence of any other forces, the cup would accelerate downwards. However, the cup is on the table, so what happens is that the gravitational force pushes down on the table, which pushes back with an exactly cancelling 'normal' force. Therefore there are two equal and opposite forces on the cup and hence the net force is zero and the cup is unaccelerated and stays in place.


    The mass of the Earth causes a curvature in the space-time around it. This curvature changes the natural 'straight line' path of objects (formally known as the geodesic) from what it would be in the abscence of curvature. This is the parth that objects with no net force (remember now that gravity is not a force in this perspective) will follow. Specifically, objects following a geodesic near the Earth will move towards the centre of the Earth (neglecting any tangential motion, lets assume our coffee cup is not in orbit). Therefore if you apply no forces to your coffee cup, it will naturally move towards the centre of the Earth, it will fall. By placing it on a table, you are now applying a force upwards. This forces continually accelerates the cup, continually preventing it from following its natural path.

    For things like coffee cups on the Earths surface, we can use either 'philosophy' and we get the same answer. Newtonian physics is easier to grasp, but it's worth trying to understand the difference between the two approaches. Once the gravitational fields get strong enough, or the coffee cup is moving fast enough, we have to use Einsteins version, because that is the only one that gets the right answer.

    The reason the Einsieinian approach may be tricky to grasp at first is that it flies in the face of the way we normally think of our everyday existance. Standing still on the ground feels like the normal, neutral state, so it's unsurprising that Newtonian physics describes this as being unaccelerated. On the other hand, freely falling is an unusual thing for us, so it feels like that is the situation in which we are being accelerated. One thing to realise though is when you are standing on the ground, you can clearly feel the acceleration upwards due to the ground, do it for long enough and you'll get sore feet! On the other hand, when you are falling you are 'weightless', in the abscence of air resistance, if you closed your eyes you wouldn't even know which way was 'down', because there are no forces being applied to you to give you a reference direction.
  9. Dec 5, 2009 #8
    Let's assume that the gravity around the Earth disappeared and instead a rocket engine should keep the moon in orbit around the earth.
    How much energy (per second) should be used?
    How can that be calculated.

    Total KE is 3E29 J according to my calculation, based on 1000 m/s.
    I guess it’s the same amount of energy that is reqiured to keep the moon obiting.

    Then divide with second per orbit to find the KE per orbit ?

    But I am confused, because when adding the twice KE the speed must increases too, but not to the double, but so far I think only 50 % - because of the resistance against increasing speed is increasing 4 times also...
  10. Dec 5, 2009 #9


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    You wouldn't need to expend 3E29J of energy per second to keep the moon in orbit. You may need that much energy to get the moon to 1000m/s, but once it's at 1000m/s that's the energy it possesses, if you added energy to it, it would speed up.

    No energy is expended to keep the moon in orbit, since both its total energy (KE+PE) do not change.

    If you replaced the Earth's gravity with a rocket, the rocket would not do work on the moon since it would need to apply the force PERPENDICULAR to the direction that the moon is moving at all times. Remember P=F dot v and in this case, F and v are always orthogonal so the power is zero.
  11. Dec 6, 2009 #10
    I mean pretend the Erath (and gravity from Earth) was not there, and the moons still should orbit like it does.
    I would so fare I can understand need twice the moons velocity KE to turn the moons motion direction, and this happens twice per orbit, - which mean total 4 times so much KE to keep the moon in orbit.?
    How much energy would it require (per second or per orbit) to keep the moon in orbit, if gravity did not exsist ?
    Last edited: Dec 6, 2009
  12. Dec 8, 2009 #11
    I am bit confused with some force is applied but no energy spent, like holding some weight in your hand and standing still on earth.

    In the moon case referred above, if some mechanism id added to change the angle of rocket firing to keep it always perpendicular to the motion, the moon will take a circular path similar to some external gravitational force acting on it. All the time the rocket is doing some work - loosing energy in the form spent fuel (additional power is also spent in continuously changing the firing angle). Does it mean then that in case of gravity, it does the work of the rocket? and how much is it ?
  13. Dec 8, 2009 #12


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    Yes, there is also no energy imparted on the object you are holding. Consider a book sitting on a table. There is a force, but no energy.

    Please understand, we are talking about output here, not input. There are some systems that need to expend energy to create force, but if there is no motion, there is no ouput energy and in a sense, 0% efficiency.
    Still zero. The rocket is expending energy and doing no work.
  14. Dec 8, 2009 #13


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    As discussed in the other thread, still zero.

    Now if you want to know how much energy is required of a rocket tasked to do this job, since there is no output work, there is no single answer: it is entirely dependent on the particulars of the rocket. Ie, the lower the exit velocity of the exhaust gases, the lower the energy required for the same force. This is why modern commercial jets use high bypass turbofans.
    Last edited: Dec 8, 2009
  15. Dec 8, 2009 #14
    1) In case of forced circular motion about a point (without gravity), the energy is spent with no output
    2) With circular motion caused by equivalent gravitational attraction from the that point, the revolving object is not spending any energy. The rocket need not fire and consequently no fuel will be spent.
    In the 2nd case as compared to case1, there is saving of fuel energy. Can this not be considered as the energy caused by gravitation?
  16. Dec 8, 2009 #15


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    No, it can't be. As you correctly pointed out, neither case 1 nor case 2 involve any energy output. The savings is 0-0=0. And as I said in the post you quoted, the energy input could be literally anything.

    Think about it this way:
    A kid says to his dad: Daddy, daddy, I saved $2 today by walking to the park today instead of taking the bus!
    The dad slaps him and says: You dope, you could have saved more by walking next to a taxi!

    So to it is with this problem. Zero energy is zero energy is zero energy. It doesn't matter if something that can be done with zero energy could also be done with a lot of energy. It is useless to say you have saved anything when you didn't really have to spend anything in the first place. Alternately, with a little twist of logic, you could claim to have saved anything you want!

    And please understand: this isn't just about orbits. Force (torque) and energy (or power) in general are just not related in the way you are trying to relate them. The same type of problem applies to your car, for example. If you were to ask "How much power is required to generate 100 ft-lb of torque", that question would also have no answer. Or how about: "How much power is required to lift a 1N book?" Also no answer.
    Last edited: Dec 9, 2009
  17. Dec 9, 2009 #16
    Acceleration due to gravity is the force that in the end is the "input" the force that causes planets and moons to orbit. And this force is not “stable”.

    This force is almost 4000 so strong near the surface of the Earth, as it is in the orbit of the moon. But the speed of a satellite would only be 8 times so fast near the surface.

    Its seems that their is a lost of speed.

    [PLAIN]http://www.science27.com/forum/a+b.jpg [Broken]

    If object A and B was fastened to a string, - and the speed of A would be twice than object B, -a Newton meter would show twice so much force for A than for B.

    This shows that an orbiting object planet / moon must create a force opposite to gravity.
    Is such “created opposite force” free ?
    I mean at least speed would decrease right ?

    If the bearing was rusty and some of the energy was converted to heat, - object A would cause twice that heat than B. – I mean what about such case now you would certainly lose energy? – or at least transfer speed, right ? – Could it be possible to say how much ?
    Last edited by a moderator: May 4, 2017
  18. Dec 9, 2009 #17
    force = m*v^2 / r, so

    force ratio = [(mA/mB)*(vA/vB)^2 ]/ (rA / rB)

    If A moves twice as fast as B, then nothing states that the force on A need be twice than on B, there are still two unspecified, fully independant degrees of freedom: the ratio of the orbital radii, and the mass ratio of A and B.

    Speed would not decrease unless you have something removing energy from the system. In circular motion, the force is perpendicular to the displacement, so the work done (dot product of these two vectors) is zero.
  19. Dec 9, 2009 #18


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    Free of energy, yes.
    That's a meaningless question - objects do not spontaneously jump from one orbit to another. And moving from one orbit to another involves a potential energy change in exactly the same way that moving in an elevator involves a potential energy change.
    Certainly, a rusty bearing causes an energy loss and causes the rotating apparatus to slow down. But that doesn't have anything to do with orbital dynamics: there are no rusty bearings in an orbit.
  20. Dec 27, 2009 #19
    When considering velocity as a function of radius for the above circular motion equations the answer gives the perfect speed needed to stay at whatever radius. If an object is moving any slower it will fall inward along some new path. likewise if it's moving faster it will fall away along some new path.
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