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ganeshie8
 one year ago
For some unavoidale circumstances, two astronauts Peter and Mary are orbiting around earth as shown. Mary realizes that she has forgotten her lunch box. Peter happens to have an extra lunch box which he decides to pass to Mary. At what velocity Peter must throw the lunch box ?
They are separated by an arc distance of 1/5th the circumference of the orbit.
ganeshie8
 one year ago
For some unavoidale circumstances, two astronauts Peter and Mary are orbiting around earth as shown. Mary realizes that she has forgotten her lunch box. Peter happens to have an extra lunch box which he decides to pass to Mary. At what velocity Peter must throw the lunch box ? They are separated by an arc distance of 1/5th the circumference of the orbit.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443419836472:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(7000km\) is the distance from the "center of earth" to Mary/Peter's orbit

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3they are orbiting the earth, so angular velocity \(\omega\) cannot be \(0\), right ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.4what im asking is are they orbitting at same angular velocity

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Oh ok, let me add that info, sorry... i must give the distance between Peter and Mary

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I'll update that in the main question quick..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2This problem seems confusing from how it's worded. When the lunchbox is thrown, it should follow a circular path.\[\frac{v^2}{r} = \frac{GM}{r^2}\tag{gravity provides the centripetal force}\]

dan815
 one year ago
Best ResponseYou've already chosen the best response.4we are going to also hae to make some assumptions like this question is trivial if u dont say that any velocity greater than their veolicty will escape earths ggravity, then the same force that is holding these 2 in orbit will hold whatever they throw in orbit as long as its veloicty is not too high

dan815
 one year ago
Best ResponseYou've already chosen the best response.4if we apply normal physics then this is really just a parabollic equation problem

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dw:1443420236862:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2We're probably gonna have to use the conservation of angular momentum here. Probably.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So it's not necessary that the lunchbox follows a circular trajectory  that right?

dan815
 one year ago
Best ResponseYou've already chosen the best response.4if u think about this problem its really the same as theese 2 ppl on another planet with higher radius on the planet

dan815
 one year ago
Best ResponseYou've already chosen the best response.4that is if dw/dt=0 meaning their distance apart is constant

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3lunch box follows whatever the trajectory that gravity decides, for sure, its not a parabola or circle

dan815
 one year ago
Best ResponseYou've already chosen the best response.4okay so just answer that 1 questio nis the distance between them constant

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes, but it depends on the angle with which the thing is thrown, right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dan, i have updated the question, please scroll up and read again. `they are separated by an arc distance of 1/5th the circumference of the orbit`

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3since they are in the same orbit, their velocities will be same, so yeah they are always at 1/5*circumference away from each other

dan815
 one year ago
Best ResponseYou've already chosen the best response.4okay so ya like parth is suggesting, there are multiple solutions based on the angle chosen

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3there is only one solution depends on what trajectory an object in orbit takes when you increase/decrease its velocity

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2so here is how I think of it: what if the guy throws it away from himself and the earth so that it goes away for some time and then starts to come back, and by the time it comes back, the other astronaut is right underneath it and catches it?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443420881110:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Eactly! dw:1443420863876:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2not tangentially but normally

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dan that can't happen, since you have increased the "speed" of the lunch box, i think the lunch box must take a "bigger" orbit

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2wait, are you talking to me?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3talking to this one : dw:1443421091159:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2ah, that's not true. when you throw a ball upwards, does it start orbiting the earth? :P it starts orbiting only when the *velocity is perpendicular* to the radius and \(GMm/r^2 = mv^2/r\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if you throw a stone horizontally with sufficient speed, such that it crosses the curvature of earth before it falls down, it will be in orbit, right ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2the orbital velocity of the astronauts is given by \(v = \frac{GM}{r}\) and the other astronaut must cover fourfifth of the orbit before the ball comes back which is \(d = \frac{8\pi r}{5}\). the time of flight of the stone will be the same as \(d/v = \frac{8\pi r^2}{5GM} = t\). now this is just a vertical projectile and we have to determine the initial velocity from a given parameter, the time of flight.

dan815
 one year ago
Best ResponseYou've already chosen the best response.4okay heres a solution that must work then i guess

dan815
 one year ago
Best ResponseYou've already chosen the best response.4how about throwing 2*Velocity backwards

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yes, but that's horizontal. we're throwing it vertically away from the earth.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2that's true. 2*velocity relative to the astronaut. it follows a clockwise orbit in that case. nice.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443421559981:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3one moment please if you give it an impulse perpendicular to the direction of Peter, would it come back to the same orbit ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think, it can never come back to the same orbit because the "speed" of an object uniquely determines the orbit radius

dan815
 one year ago
Best ResponseYou've already chosen the best response.4if u ask me its better to ignore this physics and just write out pure equations

dan815
 one year ago
Best ResponseYou've already chosen the best response.4we might fight all kinds of solutions just working with these differential equations

dan815
 one year ago
Best ResponseYou've already chosen the best response.4start with the path equation of mary we can use peter just for initial conditions

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Alright, for simplicity we may assume the period of both peter and mary is 100 minutes.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443421917404:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2no, we can actually find out the period...\[\frac{2\pi r}{ \sqrt{\frac{GM}{r}}} = \frac{2\pi r^{3/2}}{\sqrt{GM}}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443422101849:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3right, that should come around 100 minutes

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[a = \frac{dv}{dt} = v\frac{dv}{dr }\]\[a ~dr = v~ dv\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Well, that equation doesn't really help. That just tells us that the thing comes back with the same speed that it goes out with. :

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Could anyone suggest an equation to use with this?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Either we could take the acceleration to be constant, or...

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Dan, Ganeshie... anyone?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i don't know, i couldn't comprehend yet, what happens when you throw an object radially out from a satellite in orbit

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3never thought about it before, so im gonna think about it a bit..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2arrey, have you ever thrown a ball vertically up? what happens? it comes back.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3but the satellite also got tangential velocity right

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2then relative to the astronaut, he can throw it at a velocity that cancels out the tangential component

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2we're mostly interested in the normal component though

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3im still unsettled on that, but il move on. so you want to use the "flight time" of lunch box, such a way that Mary goes exactly below the lunch box and catches it ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2an equation that involves a, t, u and r. hmm. in the case of constant acceleration, that equation is \(s = ut + \frac{1}{2}at^2\) so i believe we'll have to use the calculus counterpart of that? i still feel that because \(a\) is a function of \(r\) we have to use \(a~ds = v~dv\). Argh!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Okay I see now, so the gravity eats up the normal component of velocity completely

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3tangential component will be same as that of Peter's

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2of the ball? nah, it would be thrown such that it cancels out the tangential component completely.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2though I like dan's idea better  we can just make it have the same tangential component in the opposite direction

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i was thinking you both had the same idea let me scroll up read again

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Okay so you want to kill the tangential component of velocity and give it some normal velocity radially out, nice

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2if we talk in your terms taking t = 100 minutes, we're throwing a ball upwards and it takes 4/5 * 100 minutes = 80 minutes to come back which is the same as the time Mary takes to be right underneath

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Wow, I see now, that should work perfectly, at least theoretically !

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3sorry i am just thick at learning new stuff, takes time..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yes! as long as it doesn't exceed the escape velocity, we should be fine

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you're throwing radially out dan is throwing it with an angle so that it gets a parabolic trajectory

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Dan wrote a better idea afterwards: giving it a velocity so that it starts to go in the same orbit as the astronauts, but backwards!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah 80 minutes of flight time might take the lunchbox out of earth's gravity, but it should work if the speed is less than \(\sqrt{2} v_{orb}\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2by symmetry the first half of the motion should take 40 minutes\[a dr = v dv \]\[\int_{7\cdot 10^6 }^{r}  \frac{GM}{r^2 }dr = \int_u^0 vdv \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2hmm, apparently time doesn't seem to come anywhere in the equation so far so maybe a separate one is needed for that?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2err why doesn't it work

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2which equation should i use to get the time factor in

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3can't we assume the acceleration is constant since 7000km is near earth ?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2we definitely can, but I'm not trying to leave anything out. it may go really far out you know...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3ahh 40 minutes maybe too long for flight time, we don't know yeah

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2the flight time is 80 minutes but I divided it into two halves why is it giving me so much trouble? err

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2dw:1443424701627:dw

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2why am I not able to solve an elementary motion problem?!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2oh lol\[a dr = vdv\]\[GM \int_{r_0}^r \frac{1}{r^2 } \cdot dr = \int_{u}^0 v \cdot dv\]\[2GM\left(\frac{1}{r_0}  \frac{1}{r }\right) =u^2 \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2now we need to determine what \(r\) is from the time...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3can we use work done = change in kinetic energy power = dw/dt

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2wow lol, the equation I just found says exactly that, decrease in kinetic energy = increase in potential energy

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3yeah need to use that power equation somehow because that is where time comes into the picture

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2this problem should be easier to solve than what it is right now let me ask my uncle for a moment

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Okay I don't have any clue how to use 40 minutes.. I'll wait :)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2me neither  that's why I'm confused. and I'm sure that it's not overinformation because we can always increase or decrease \(u\)  it's a variable. so far, we've treated it as a constant term.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2@vishweshshrimali5 is also online in my friends' list. do you know him?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2oh my, what do you think about double integration?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you want to factor in both \(r\) and \(t\) at the same time im useless here, never did these problems before

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Then your next project is kinematics.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2to do this fully i reckon requires consideration of stuff like this [ripped straight from wiki]: \[\ddot{\boldsymbol{r}} = \left( \ddot r  r\dot\varphi^2 \right) \hat{\mathbf r} + \left( r\ddot\varphi + 2\dot r \dot\varphi \right) \hat{\boldsymbol{\varphi}} \ = (\ddot r  r\dot\varphi^2)\hat{\mathbf{r}} + \frac{1}{r}\; \frac{d}{dt} \left(r^2\dot\varphi\right) \hat{\boldsymbol{\varphi}}\] so in the radial direction we have \(\large \ddot r  r\dot\varphi^2 = \frac{GMm}{r^2}\) tangentially we have \(\large \ddot\varphi + 2\dot r \dot\varphi=0\) but as Wiki shows we can also say that \( \frac{1}{r} \frac{d}{dt} \left(r^2\dot\varphi \right) = 0\) so \(\large r^2\dot\varphi = const \) which i think leaves you to solve this: \(\large \ddot r = \dfrac{K}{r^3}\dfrac{GMm}{r^2} \) ....horrible the back of a owlet packet is to create your own g in space as we do on earth which means \(g = \frac{GM}{r^2}\) and use simple equations of motion , here \(x = t(u + \frac{1}{2}at)\) with zeros at \(t=0 , \frac{2u}{a} \) , ie throwing the thing radially outwards and waiting for it to come back we know that a fifth period is \(T_{1/5} = \frac{2 \pi}{5}\sqrt{\frac{r^3}{GM}}\) so \(\frac{2u}{a} =2u\frac{r^2}{GM} =\frac{2 \pi}{5}\sqrt{\frac{r^3}{GM}}\) \[u = \frac{\pi}{5} \sqrt{\frac{GM}{r}} \approx 4.74m/s\] the escape velocity is \[v_e = \sqrt{\frac{2GM}{r}}\] so that should work i think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @ParthKohli wow lol, the equation I just found says exactly that, decrease in kinetic energy = increase in potential energy \(\color{#0cbb34}{\text{End of Quote}}\) This is one BIG concept about this problem. *ding ding ding*

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2can anyone help me finish the problem now that this thing is getting the attention it deserves?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0YESSS!!! I would use escape velocity to solve this problem as well, also according to what dan said in the beginning about it. Sorry, I was reading through the thread :P

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I'm going to repeat what I'm asking.dw:1443434874247:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @dan815 we are going to also hae to make some assumptions like this question is trivial if u dont say that any velocity greater than their veolicty will escape earths ggravity, then the same force that is holding these 2 in orbit will hold whatever they throw in orbit as long as its veloicty is not too high \(\color{#0cbb34}{\text{End of Quote}}\) This portion I believe

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Solve for \(u\) if the parameters given are:\[a(r) = \frac{GM}{r}\]\[t = 80\cdot 60 ~\rm sec\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Are there any constraints on the initial and final velocity of the lunch box? If not, throwing it as close to the speed of light as possible would certainly be fastest.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3The chances are you can throw it with any velocity as long as you aim in the right direction.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3right, we need to choose the correct velocity of lunch box... WIth wrong velocity, it misses Mary, and it will be in orbit; depending on the LCM of periods, Mary might be able to catch it after few revolutions...

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2i think peter must throw the box with a velocity \[2\sqrt{\frac{ GM }{ r }}\]dw:1443436931036:dw this will make the box follow that circular path nd hence it wuld prolly collide with mary

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Suppose the trajectory of the lunch box is described by s(t). We require \(v\left(t_0\right)=v\left(t_1\right)=0\). \[ v(t)=\frac{ds}{dt}\\ \frac{1}{v(t)}=\frac{1}{ds/dt}\\ \int_{t_0}^{t_1}\frac{1}{v(t)}\,ds=\int_{t_0}^{t_1}\frac{1}{ds/dt}\,ds=t_1t_0\\ \int_{t_0}^{t_1}\frac{1}{v(t)}\,ds=\int_{t_0}^{t_1}\frac{1}{v(t)}\,\sqrt{r(t)^2+\left(\frac{dr}{d\theta}\frac{d\theta}{dt}\right)^2}\,dt\\ \] Is that correct?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think our goal here is to find \(v(t_0)\), why do we require it to be 0 ?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3No reason except Mary has to catch it. I think if \(v\left(t_0\right),v\left(t_1\right)\neq0\) there is no unique solution.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3Lagrangian mechanics? \[ L=TV\quad \text{T is total kinetic energy, V is total potential.}\\ L=\frac{1}{2}mv^2+\frac{Gm}{r}=0\text{ by energy conservation.}\\ \frac{\partial L}{\partial \mathbf{r}}  \frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf{r}}} =0\\ r=\\mathbf{r}(t)\\\ v^2=\mathbf{v}(t)\cdot \mathbf{v}(t)=\mathbf{\dot{r}}(t)\cdot\mathbf{\dot{r}}(t) \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3@imqwerty \(v_{esc}\) is \(\sqrt{\dfrac{2GM}{r}}\), Is your expression related to this ? I think ur direction should work as it reduces the speed, there by reducing the orbit period of lunchbox : dw:1443438211798:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Interesting, whats the trajectory of lunchbox that you're assuming @thomas5267

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3That is the equation that yields the trajectory if I am not mistaken.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.3The partial derivatives with respect to the vectors are componentwise partial derivatives. https://en.m.wikipedia.org/wiki/Lagrangian_mechanics#Equations_of_motion

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2that lagrangian will lead you back to the acceleration equation listed above, except it will be deficient in that it lacks the tangential term. even leaving aside gravity, as the thing moves further away from earth it's angular velocity must fall in order to preserve momentum/ energy. hence the \(\large \ddot\varphi + 2\dot r \dot\varphi=0\) terms that follows.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes @ganeshie8 and @thomas are u throwing the box like thisdw:1443438908909:dw