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#1
Nov1205, 01:05 PM

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For an object put in to position say halfway to the moon.
Take away all orbital and any other speed – that is it’s completely stationary with respect to the earth. Of course it will immediately start to freefall directly to earth. Acceleration is increasing so calculating the speed at different points cannot use a simple acceleration formula. Is there a straight forward formula that gives its speed as a function of distance that accounts for the increasing acceleration? RB 


#2
Nov1205, 01:24 PM

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Halfway to the moon, the moon's gravity is only 1/80 th of the Earth's gravity, so it is still reasonable to negelect the moon's gravity. The easy way to solve the problem of an orbit around a single body is too note that there are two conserved quantites, energy and angular momentum. If the angular momentum is zero, as it would be for a body falling straight in to the Earth, the problem is even easier Energy = m(GM/r+ .5 m v_r^2 + .5 m v_theta^2) Angular Momentum = m r v_theta The fact that angular momentum is zero means v_theta is zero, thus the velocity of the object at every point is strictly radial (directly towards the planet) and the solution of the first equation gives v = v_r as a function of r. If for some reason the angular momentum isn't zero then you need to fingure out v_theta first from the value of Angular momentum, and then compute v_r from the first formula. The total velocity v will be v= sqrt(v_r^2 + v_theta^2). 


#3
Nov1205, 03:59 PM

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v = v_r I don’t understand that – maybe it’s the font’s Or I just don’t know how to read it Is r radius ?? What is v_ ? If V_{max} = Max speed When object were allowed to fall to center of earth core & all that mass at that single center point Do you mean r = (r_{max}  r_{current} ) So r would be zero when object at limit and at a max when object crosses center core of earth? Then “v_” = V_{max}/ r_{max} But that seems to liner. I must be missing something in the text fonts. 


#4
Nov1205, 05:52 PM

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Free Fall Speed
Okay,
The total energy of the object can be expressed as [tex]E_t = \frac{mv^2}{2}  \frac{GMm}{r}[/tex] m is the mass of the object M the mass of the Earth v the velocity of the obect r is the distance from the center of the Earth. At its start the object's velocity is zero, so we get [tex]E_t = \frac{GMm}{r_{int}}[/tex] where R_{int} is the initial starting distance from the Earth. Energy is conserved so [tex]  \frac{GMm}{r_{int}} = \frac{mv^2}{2} \frac{GMm}{r}[/tex] [tex]  \frac{GM}{r_{int}} = \frac{v^2}{2} \frac{GM}{r}[/tex] [tex] \frac{GM}{r}\frac{GM}{r_{int}} = \frac{v^2}{2}[/tex] [tex] 2GM \left(\frac{1}{r}\frac{1}{r_{int}} \right) = v^2[/tex] [tex] \sqrt{2GM \left(\frac{1}{r}\frac{1}{r_{int}} \right)} = v[/tex] Since the new radius r is equal to the distance traveled minus the distance traveled: [tex]r= r_{int}d[/tex] then: [tex] \sqrt{2GM \left(\frac{1}{r_{int}d}\frac{1}{r_{int}} \right)} = v[/tex] which gives you the velocity after falling a distance d from a height of r_{int} 


#5
Nov1205, 10:30 PM

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v_r is the radial component of the velocity, the component directed towards the Earth.
v_theta is the angular component of the velocity, perpendicular to v_r. Earth.........x at point 'x', v_r points to the left, v_theta points up If I understand your question correctly, v_theta is zero, and thus v=v_r. You then just substitute v=v_r in the first equation (the second equation becomes irrelevant) to find v as a function of r. Janus's post was much more complete, hopefully he explains it all in more detail and more clearly than I did. 


#6
Nov1305, 02:00 PM

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The implication is the speed at the earth center should go to the sqrt of infinity. I think your working with conserving the energy as measured from the distant infinity. That is the energy being conserved gets smaller at greater r. Thus it looks like your solving for the escape speed which does get smaller at greater r. Here were looking at an increasing potential as the farther out we start the more energy we have to convert to higher speeds as we approach the center of the gravitational sorce. Maybe I need to work on replacing the g with GM/r^{2} in the gravity formula somehow solving for r or d. 


#7
Nov1305, 04:33 PM

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The formula is good only when the object never gets closer to the center of the Earth than the Earth's radius  i.e. that objects stop when they hit the Earth's surface.
If we let r_int = infinity, then the formula reduces to the formula for escape velocity as per the wikipedia http://en.wikipedia.org/wiki/Escape_velocity (with the appropriate change of variable). You might also want to review the wikipedia (or a physics text) about gravitational potential energy  the formula follows directly and simply from the conservation of energy. http://en.wikipedia.org/wiki/Gravitational_potential 


#8
Nov1305, 04:43 PM

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[itex]\frac{Mmv^2}{2}[/itex] is the kinetic energy and [itex] \frac{GMm}{r}[/itex] is the gravitational potential energy Note.as r increases, the equation becomes less negative and the potential energy increases. [quote] [tex]E_t = \frac{Mmv^2}{2} \frac{GMm}{r}[/tex] Simply means that the total energy is equal to the kinetic energy plus the gravitation potential energy. 


#9
Nov1405, 11:27 AM

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I think is see what your doing, your calculating the escape speeds for both r_{start} and r_{end}; then using Pythagorean theorem to find the diff. But that has limits: thus you can’t find answers when you get close to the center. Example: for the Sun figuring the escape speed for a spot about 1 km form the point source center. Using GM = 132 x 10^{9}. You get a fall from infinity speed or an escape speed of about 360,000 km/s. Obviously we are inside Swartzchild radius. Not very useful at these limits without taking relativity into account. But if we limit our drops to inside the solar system, just like we need to limit the relative masses so that m<<<M, we should be able to create a “straightforward” formula that can give those speeds including ‘the center’ where r=0 for the max speed befor slowing down starts, after passing the center. In orbitial terms I guess the center would become the perihelion for the "striaight line orbit" with no angular movement. NOTE: A good devise for figuring escape velocities using conservation. That’s not what we are doing here. We are using the real potential energy from the start spot, back down to the center. And should be able to get speeds all the way down to & including r = 0. (Limited to cases where the speed remains well lower than c.) It’s so standard to use the escape velocity assumptions, must be why I’m have trouble coming up with the proper way to derive this without using the familiar “negative gravitational potential energy” shortcut to get a more direct straightforward formula that dosn't need the Pythagorean theorem. 


#10
Nov1405, 03:33 PM

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#11
Nov1405, 04:13 PM

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is pretty straight forward. It's basically Leibniz's visviva equation. If you substituted the semimajor axis for r_int, you could use it to find the speed of an object in an elliptical orbit, as well. You are right that it could be used all the way down to r=0, provided all the mass really did exist in a point. Once you're below the surface, some of the mass is above you and the GM part will no longer be the same. 


#12
Nov1405, 07:53 PM

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Is the square of the starting point speed. [tex] {2GM \left(\frac{1}{r} \right)}[/tex] Is the square of the ending point escape speed. (It’s this ending point that is a problem as r nears 0.) Combining the squares ; in this case taking the diff. since the E is being measured as a negative from infinity. And then taking a SQRT of the sum is basic Pythagoream … 


#13
Nov1505, 12:00 AM

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Relativistic corrections would be necessary for the same calculation done to a black hole or neutron star. [tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex] if you like. It will give you the same answer. 


#14
Nov1505, 09:00 AM

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Given a start position of r = 100r_{earth} A start velocity of v = 0 Defining distance d = 100r r What is your solution for v as a function of d ?? EXAMPLE on the surface of the earth I can use V = 8 ft/sec * sqrt d At 16 feet that gives me 32 feet per sec Good answer formula comes from acceleration a=32 feet per sec per sec but it wont last long as a is getting stronger as d gets longer. So for a complete straight forward formula with a known starting factor of 8ft/s V = (A*8ft/s +B) * sqrt d + C Where A B and C are each some function of d or a constant like 1 or 0. I just don’t know how to get to the simple formula directly calculating from d so that can be solved for d = starting radius. So far the solutions I’ve seen strike me as incomplete integrations that havn’t accounted for taking the limit of d to it’s max of starting r. so that for d greater than starting r ; v would be decreasing from that max speed found at r=0. 


#15
Nov1505, 02:15 PM

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[tex] \sqrt{2GM \left(\frac{1}{r}\frac{1}{r_{int}} \right)} = v[/tex] r_{int} = 100 r_{earth} BTW, suspect you really meant that d = 100 r_{earth} r This gives [tex] v = \sqrt {2}\sqrt {GM \left( \left( 100\,{\it re}d \right) ^{1}{ \frac {1}{100}}\,{{\it re}}^{1} \right) } [/tex] Note that a = dv(d)/dd * dd/dt = dv/dd*v = GM / (100 r_earth  d)^2 Note also that the answer, and the differential equation you started with, are both incorrect for r<r_{earth}, i.e for d > 99 r_{earth} You will need to use the correct force law for a body inside the Earth if you really want to solve for the motion there. 


#16
Nov1505, 03:56 PM

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As assumed before the mass is a point source. The point is to have a formula capable of solving for small r even r=0. Compare μ for the sun is much larger than μ for earth. (μ = GM). Thus we should expect that the Theoretical Maximum speed at crossing the center of the core point mass (without hitting it, it’s just a theoretical solution) to be much greater for the sun. But these will not solve that Theoretical Max, as they blow up at r = 0. And of course the Max Speed, obviously must occur at r= 0. There is nothing in the formulas that demand the Mass (earth) have some fixed radius that the falling object must hit. With the mass as point source your right these formats of the solution are likely just starting from the wrong place to be able to work that solution. I just don’t know the correct place to start from. Yes, if the mass is given a fixed diameter with uniform density or at least a changing density with a reasonable function to r , it would need adjustments to the solution I’m looking for here. This of course assumes the falling object is “transparent” to the mass (or at least someone has dug an appropriate hole. You describe those adjustment as the “correct force law for a body inside the Earth”; is there someplace that describes doing this. Maybe some clues there. Thanks 


#17
Nov1505, 04:59 PM

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A funny thing about Physics, it works best when you keep the problems physical. Your requirement is non physical therefore you can expect non physical results. Real planets have a radius, your initial problem statement was for an object between the earth and the moon, both the earth and moon have a radius. You have been given a solution to that problem. 


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