ganeshie8
  • ganeshie8
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
|dw:1443419836472:dw|
ganeshie8
  • ganeshie8
\(7000km\) is the distance from the "center of earth" to Mary/Peter's orbit
dan815
  • dan815
|dw:1443419927131:dw|

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ganeshie8
  • ganeshie8
they are orbiting the earth, so angular velocity \(\omega\) cannot be \(0\), right ?
dan815
  • dan815
change in w
dan815
  • dan815
what im asking is are they orbitting at same angular velocity
ganeshie8
  • ganeshie8
Oh ok, let me add that info, sorry... i must give the distance between Peter and Mary
ganeshie8
  • ganeshie8
I'll update that in the main question quick..
ParthKohli
  • ParthKohli
This 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
  • 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
dan815
  • dan815
if we apply normal physics then this is really just a parabollic equation problem
dan815
  • dan815
|dw:1443420214293:dw|
ganeshie8
  • ganeshie8
|dw:1443420236862:dw|
ParthKohli
  • ParthKohli
We're probably gonna have to use the conservation of angular momentum here. Probably.
dan815
  • dan815
|dw:1443420284948:dw|
ParthKohli
  • ParthKohli
So it's not necessary that the lunch-box follows a circular trajectory - that right?
dan815
  • dan815
if u think about this problem its really the same as theese 2 ppl on another planet with higher radius on the planet
dan815
  • dan815
that is if dw/dt=0 meaning their distance apart is constant
ganeshie8
  • ganeshie8
lunch box follows whatever the trajectory that gravity decides, for sure, its not a parabola or circle
dan815
  • dan815
okay so just answer that 1 questio nis the distance between them constant
ParthKohli
  • ParthKohli
Yes, but it depends on the angle with which the thing is thrown, right?
ganeshie8
  • ganeshie8
dan, 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
  • ganeshie8
since 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
  • dan815
okay so ya like parth is suggesting, there are multiple solutions based on the angle chosen
ganeshie8
  • ganeshie8
there is only one solution depends on what trajectory an object in orbit takes when you increase/decrease its velocity
ParthKohli
  • ParthKohli
so 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
  • ParthKohli
|dw:1443420881110:dw|
ganeshie8
  • ganeshie8
Eactly! |dw:1443420863876:dw|
ParthKohli
  • ParthKohli
not tangentially but normally
ganeshie8
  • ganeshie8
dan 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
  • ParthKohli
wait, are you talking to me?
ganeshie8
  • ganeshie8
talking to this one : |dw:1443421091159:dw|
ParthKohli
  • ParthKohli
ah, 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
  • ganeshie8
if 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
  • ParthKohli
the orbital velocity of the astronauts is given by \(v = \frac{GM}{r}\) and the other astronaut must cover four-fifth 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
  • dan815
okay heres a solution that must work then i guess
dan815
  • dan815
how about throwing 2*Velocity backwards
ParthKohli
  • ParthKohli
yes, but that's horizontal. we're throwing it vertically away from the earth.
ParthKohli
  • ParthKohli
that's true. 2*velocity relative to the astronaut. it follows a clockwise orbit in that case. nice.
dan815
  • dan815
|dw:1443421483568:dw|
ParthKohli
  • ParthKohli
|dw:1443421559981:dw|
ganeshie8
  • ganeshie8
one moment please if you give it an impulse perpendicular to the direction of Peter, would it come back to the same orbit ?
ganeshie8
  • ganeshie8
I think, it can never come back to the same orbit because the "speed" of an object uniquely determines the orbit radius
dan815
  • dan815
if u ask me its better to ignore this physics and just write out pure equations
dan815
  • dan815
we might fight all kinds of solutions just working with these differential equations
dan815
  • dan815
start with the path equation of mary we can use peter just for initial conditions
ganeshie8
  • ganeshie8
Alright, for simplicity we may assume the period of both peter and mary is 100 minutes.
ParthKohli
  • ParthKohli
|dw:1443421917404:dw|
ParthKohli
  • ParthKohli
no, we can actually find out the period...\[\frac{2\pi r}{ \sqrt{\frac{GM}{r}}} = \frac{2\pi r^{3/2}}{\sqrt{GM}}\]
ParthKohli
  • ParthKohli
|dw:1443422101849:dw|
ganeshie8
  • ganeshie8
right, that should come around 100 minutes
ParthKohli
  • ParthKohli
\[a = \frac{dv}{dt} = v\frac{dv}{dr }\]\[a ~dr = v~ dv\]
ParthKohli
  • ParthKohli
Well, 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
  • ParthKohli
Could anyone suggest an equation to use with this?
ParthKohli
  • ParthKohli
Either we could take the acceleration to be constant, or...
ParthKohli
  • ParthKohli
Dan, Ganeshie... anyone?
ganeshie8
  • ganeshie8
i don't know, i couldn't comprehend yet, what happens when you throw an object radially out from a satellite in orbit
ganeshie8
  • ganeshie8
never thought about it before, so im gonna think about it a bit..
ParthKohli
  • ParthKohli
arrey, have you ever thrown a ball vertically up? what happens? it comes back.
ganeshie8
  • ganeshie8
but the satellite also got tangential velocity right
ParthKohli
  • ParthKohli
then relative to the astronaut, he can throw it at a velocity that cancels out the tangential component
ParthKohli
  • ParthKohli
we're mostly interested in the normal component though
ganeshie8
  • ganeshie8
im 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
  • ParthKohli
an 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
  • ganeshie8
Okay I see now, so the gravity eats up the normal component of velocity completely
ganeshie8
  • ganeshie8
tangential component will be same as that of Peter's
ParthKohli
  • ParthKohli
of the ball? nah, it would be thrown such that it cancels out the tangential component completely.
ParthKohli
  • ParthKohli
though I like dan's idea better - we can just make it have the same tangential component in the opposite direction
ganeshie8
  • ganeshie8
i was thinking you both had the same idea let me scroll up read again
ganeshie8
  • ganeshie8
Okay so you want to kill the tangential component of velocity and give it some normal velocity radially out, nice
ParthKohli
  • ParthKohli
if 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
  • ganeshie8
Wow, I see now, that should work perfectly, at least theoretically !
ganeshie8
  • ganeshie8
sorry i am just thick at learning new stuff, takes time..
ParthKohli
  • ParthKohli
yes! as long as it doesn't exceed the escape velocity, we should be fine
ganeshie8
  • ganeshie8
you're throwing radially out dan is throwing it with an angle so that it gets a parabolic trajectory
ParthKohli
  • ParthKohli
Dan wrote a better idea afterwards: giving it a velocity so that it starts to go in the same orbit as the astronauts, but backwards!
ParthKohli
  • ParthKohli
get out
ganeshie8
  • ganeshie8
yeah 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
  • ParthKohli
by 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
  • ParthKohli
hmm, apparently time doesn't seem to come anywhere in the equation so far so maybe a separate one is needed for that?
ParthKohli
  • ParthKohli
err why doesn't it work
ParthKohli
  • ParthKohli
which equation should i use to get the time factor in
ganeshie8
  • ganeshie8
can't we assume the acceleration is constant since 7000km is near earth ?
ParthKohli
  • ParthKohli
we definitely can, but I'm not trying to leave anything out. it may go really far out you know...
ganeshie8
  • ganeshie8
ahh 40 minutes maybe too long for flight time, we don't know yeah
ParthKohli
  • ParthKohli
the flight time is 80 minutes but I divided it into two halves why is it giving me so much trouble? err
ParthKohli
  • ParthKohli
|dw:1443424701627:dw|
ParthKohli
  • ParthKohli
why am I not able to solve an elementary motion problem?!
ParthKohli
  • ParthKohli
oh 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
  • ParthKohli
now we need to determine what \(r\) is from the time...
ganeshie8
  • ganeshie8
can we use work done = change in kinetic energy power = dw/dt
ParthKohli
  • ParthKohli
wow lol, the equation I just found says exactly that, decrease in kinetic energy = increase in potential energy
ganeshie8
  • ganeshie8
yeah need to use that power equation somehow because that is where time comes into the picture
ParthKohli
  • ParthKohli
this problem should be easier to solve than what it is right now let me ask my uncle for a moment
ganeshie8
  • ganeshie8
Okay I don't have any clue how to use 40 minutes.. I'll wait :)
ParthKohli
  • ParthKohli
me 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
  • ParthKohli
@vishweshshrimali5 is also online in my friends' list. do you know him?
ParthKohli
  • ParthKohli
oh my, what do you think about double integration?
ganeshie8
  • ganeshie8
you want to factor in both \(r\) and \(t\) at the same time im useless here, never did these problems before
ParthKohli
  • ParthKohli
Then your next project is kinematics.
IrishBoy123
  • IrishBoy123
to 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
Jhannybean
  • Jhannybean
\(\color{#0cbb34}{\text{Originally Posted by}}\) @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
  • ParthKohli
can anyone help me finish the problem now that this thing is getting the attention it deserves?
Jhannybean
  • Jhannybean
YESSS!!! 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
  • ParthKohli
I'm going to repeat what I'm asking.|dw:1443434874247:dw|
Jhannybean
  • Jhannybean
\(\color{#0cbb34}{\text{Originally Posted by}}\) @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
  • ParthKohli
Solve for \(u\) if the parameters given are:\[a(r) = -\frac{GM}{r}\]\[t = 80\cdot 60 ~\rm sec\]
ParthKohli
  • ParthKohli
@imqwerty ^^^
thomas5267
  • thomas5267
Are 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
  • thomas5267
The chances are you can throw it with any velocity as long as you aim in the right direction.
ganeshie8
  • ganeshie8
right, 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
  • imqwerty
i 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
  • thomas5267
Suppose the trajectory of the lunch box is described by s(t). We require \(v\left(t_0\right)=v\left(t_1\right)=0\). \[ v(t)=\frac{ds}{dt}\\ \frac{1}{v(t)}=\frac{1}{ds/dt}\\ \int_{t_0}^{t_1}\frac{1}{v(t)}\,ds=\int_{t_0}^{t_1}\frac{1}{ds/dt}\,ds=t_1-t_0\\ \int_{t_0}^{t_1}\frac{1}{v(t)}\,ds=\int_{t_0}^{t_1}\frac{1}{v(t)}\,\sqrt{r(t)^2+\left(\frac{dr}{d\theta}\frac{d\theta}{dt}\right)^2}\,dt\\ \] Is that correct?
ganeshie8
  • ganeshie8
I think our goal here is to find \(v(t_0)\), why do we require it to be 0 ?
thomas5267
  • thomas5267
No 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
  • thomas5267
Lagrangian mechanics? \[ L=T-V\quad \text{T is total kinetic energy, V is total potential.}\\ L=\frac{1}{2}mv^2+\frac{Gm}{r}=0\text{ by energy conservation.}\\ \frac{\partial L}{\partial \mathbf{r}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf{r}}} =0\\ r=\|\mathbf{r}(t)\|\\ v^2=\mathbf{v}(t)\cdot \mathbf{v}(t)=\mathbf{\dot{r}}(t)\cdot\mathbf{\dot{r}}(t) \]
ganeshie8
  • ganeshie8
@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
  • ganeshie8
Interesting, whats the trajectory of lunchbox that you're assuming @thomas5267
thomas5267
  • thomas5267
That is the equation that yields the trajectory if I am not mistaken.
thomas5267
  • thomas5267
The partial derivatives with respect to the vectors are componentwise partial derivatives. https://en.m.wikipedia.org/wiki/Lagrangian_mechanics#Equations_of_motion
IrishBoy123
  • IrishBoy123
that 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
  • imqwerty
yes @ganeshie8 and @thomas are u throwing the box like this-|dw:1443438908909:dw|
thomas5267
  • thomas5267
How do we codify the constraint that \(\mathbf{s_1}(0)=(7000,0),\,\mathbf{s_2}(0)=\left(7000\cos\left(\frac{2\pi}{5}\right),7000\sin\left(\frac{2\pi}{5}\right)\right)\) in terms of Lagrangian multipliers?
thomas5267
  • thomas5267
Suppose the lunch box is thrown directly opposite to the gravitaional force. \[ F=ma\\ -\frac{GMm}{r^2}=ma\\ -\frac{GM}{r^2}=a\\ -\frac{GM}{r^2}=\frac{d^2r}{dt}\\ r(0)=7000\\ r\left(\frac{4}{5}P\right)=7000 \]
thomas5267
  • thomas5267
I mean \(\dfrac{d^2r}{dt^2}\) not \(\dfrac{d^2r}{dt}\).
thomas5267
  • thomas5267
\[ v=\sqrt{\frac{GM}{r}}\\ P=2\pi r/v\\ \begin{align*} -\frac{GM}{r^2}&=\frac{d^2r}{dt^2}\\ \iint dt^2&=-\iint \frac{r^2}{GM}\,dr^2\\ \frac{1}{2}t^2+ct+d&=-\frac{1}{12}r^4\\ t&=0\implies d=-\frac{1}{12GM}(7000000)^4\\ t&=\frac{4}{5}P\implies\\ \frac{1}{2}\left(\frac{4}{5}P\right)^2+\frac{4c}{5}P-\frac{1}{12GM}(7000000)^4&=-\frac{1}{12GM}(7000000)^4\\ \frac{2}{5}P+c&=0\\ c&=-\frac{2}{5}P \end{align*} \] On street. Hopefully that's right.
thomas5267
  • thomas5267
\[ \frac{1}{2}t^2-\frac{2}{5}Pt-\frac{7000000^4}{12GM}=-\frac{1}{12}r(t)^4\\ -6t^2+\frac{24}{5}Pt+\frac{7000000^4}{GM}=r(t)^4 \] Take the quartic root?
thomas5267
  • thomas5267
What I did above was wrong since you cannot separate the variables like a first order differential equation. \[ \begin{align*} -\frac{GM}{r^2}&=\frac{d^2r}{dt^2}\\ -2\frac{dr}{dt}\frac{GM}{r^2}&=2\frac{dr}{dt}{\frac{d^2r}{dt^2}}\\ -2\frac{dr}{dt}\frac{GM}{r^2}&=\frac{d}{dt}\left(\frac{dr}{dt}\right)^2\\ \frac{2GM}{r}+C&=\left(\frac{dr}{dt}\right)^2\\ \sqrt{\frac{2GM}{r}+C}&=\frac{dr}{dt}\\ t&=\int\frac{1}{\sqrt{2GM/r+C}}\,dr\\ u=2GM/r+C\\ du=-2GM/r^2\,dr\\ r=\frac{2GM}{u-C}\\ r^2=\frac{4G^2M^2}{(u-C)^2}\\ t&=\int\frac{1}{\sqrt{2GM/r+C}}\frac{r^2}{r^2}\,dr\\ t&=-2GM\int\frac{1}{\sqrt{u}}\frac{1}{(u-C)^2}\,du\\ \end{align*} \] How do I integrate this? Wolfram Alpha returns a closed form answer. http://www.wolframalpha.com/input/?i=Integrate[1%2FSqrt[2GM%2Fr%2BC]%2Cr]&dataset=
thomas5267
  • thomas5267
Can't paste the url. Type Integrate[1/Sqrt[2GM/r+C],r] into Wolfram Alpha.
IrishBoy123
  • IrishBoy123
thomas that sleight of hand is hugely impressive but it has a more traditional form starting from the simplified \(m \ddot r = -\dfrac{GMm}{r^2} \) [ie looking purely at solutions that involve the sandwich being thrown radially away from earth and falling back down on the other astronaut ] we can say \( \ddot r = \dfrac{ d \dot r}{dt} = \dfrac{d \dot r}{dr}\dfrac{d r}{dt} = \dot r \dfrac{d \dot r}{dr}\) [the old fashioned way] so \(\dot r \dfrac{d \dot r}{dr} = -\dfrac{GM}{r^2} \) \(\dfrac{\dot r^2 }{2} = \dfrac{GM}{r}+C \) \(\dfrac{ \dot r}{\sqrt{2} } = \sqrt{\dfrac{GM}{r}+C}\) that's got any number of solutions, as i think you agree
thomas5267
  • thomas5267
That's not sleigh of hands but pure incompetence from a person who is going to studying first year undergraduate soon! I found that integration method directly from wikipedia! Having infinite amount of solution seems strange to me though. There should be a point where the lunch box reaches escape velocity and leaves the Earth's gravitation influence and starving Mary!

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