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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.

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  1. ganeshie8
    • one year ago
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    |dw:1443419836472:dw|

  2. ganeshie8
    • one year ago
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    \(7000km\) is the distance from the "center of earth" to Mary/Peter's orbit

  3. dan815
    • one year ago
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    |dw:1443419927131:dw|

  4. ganeshie8
    • one year ago
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    they are orbiting the earth, so angular velocity \(\omega\) cannot be \(0\), right ?

  5. dan815
    • one year ago
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    change in w

  6. dan815
    • one year ago
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    what im asking is are they orbitting at same angular velocity

  7. ganeshie8
    • one year ago
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    Oh ok, let me add that info, sorry... i must give the distance between Peter and Mary

  8. ganeshie8
    • one year ago
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    I'll update that in the main question quick..

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

  10. dan815
    • one year ago
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    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

  11. dan815
    • one year ago
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    if we apply normal physics then this is really just a parabollic equation problem

  12. dan815
    • one year ago
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    |dw:1443420214293:dw|

  13. ganeshie8
    • one year ago
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    |dw:1443420236862:dw|

  14. ParthKohli
    • one year ago
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    We're probably gonna have to use the conservation of angular momentum here. Probably.

  15. dan815
    • one year ago
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    |dw:1443420284948:dw|

  16. ParthKohli
    • one year ago
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    So it's not necessary that the lunch-box follows a circular trajectory - that right?

  17. dan815
    • one year ago
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    if u think about this problem its really the same as theese 2 ppl on another planet with higher radius on the planet

  18. dan815
    • one year ago
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    that is if dw/dt=0 meaning their distance apart is constant

  19. ganeshie8
    • one year ago
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    lunch box follows whatever the trajectory that gravity decides, for sure, its not a parabola or circle

  20. dan815
    • one year ago
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    okay so just answer that 1 questio nis the distance between them constant

  21. ParthKohli
    • one year ago
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    Yes, but it depends on the angle with which the thing is thrown, right?

  22. ganeshie8
    • one year ago
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    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`

  23. ganeshie8
    • one year ago
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    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

  24. dan815
    • one year ago
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    okay so ya like parth is suggesting, there are multiple solutions based on the angle chosen

  25. ganeshie8
    • one year ago
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    there is only one solution depends on what trajectory an object in orbit takes when you increase/decrease its velocity

  26. ParthKohli
    • one year ago
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    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?

  27. ParthKohli
    • one year ago
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    |dw:1443420881110:dw|

  28. ganeshie8
    • one year ago
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    Eactly! |dw:1443420863876:dw|

  29. ParthKohli
    • one year ago
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    not tangentially but normally

  30. ganeshie8
    • one year ago
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    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

  31. ParthKohli
    • one year ago
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    wait, are you talking to me?

  32. ganeshie8
    • one year ago
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    talking to this one : |dw:1443421091159:dw|

  33. ParthKohli
    • one year ago
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    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 \(GMm/r^2 = mv^2/r\)

  34. ganeshie8
    • one year ago
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    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 ?

  35. ParthKohli
    • one year ago
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    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.

  36. dan815
    • one year ago
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    okay heres a solution that must work then i guess

  37. dan815
    • one year ago
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    how about throwing 2*Velocity backwards

  38. ParthKohli
    • one year ago
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    yes, but that's horizontal. we're throwing it vertically away from the earth.

  39. ParthKohli
    • one year ago
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    that's true. 2*velocity relative to the astronaut. it follows a clockwise orbit in that case. nice.

  40. dan815
    • one year ago
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    |dw:1443421483568:dw|

  41. ParthKohli
    • one year ago
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    |dw:1443421559981:dw|

  42. ganeshie8
    • one year ago
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    one moment please if you give it an impulse perpendicular to the direction of Peter, would it come back to the same orbit ?

  43. ganeshie8
    • one year ago
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    I think, it can never come back to the same orbit because the "speed" of an object uniquely determines the orbit radius

  44. dan815
    • one year ago
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    if u ask me its better to ignore this physics and just write out pure equations

  45. dan815
    • one year ago
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    we might fight all kinds of solutions just working with these differential equations

  46. dan815
    • one year ago
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    start with the path equation of mary we can use peter just for initial conditions

  47. ganeshie8
    • one year ago
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    Alright, for simplicity we may assume the period of both peter and mary is 100 minutes.

  48. ParthKohli
    • one year ago
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    |dw:1443421917404:dw|

  49. ParthKohli
    • one year ago
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    no, we can actually find out the period...\[\frac{2\pi r}{ \sqrt{\frac{GM}{r}}} = \frac{2\pi r^{3/2}}{\sqrt{GM}}\]

  50. ParthKohli
    • one year ago
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    |dw:1443422101849:dw|

  51. ganeshie8
    • one year ago
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    right, that should come around 100 minutes

  52. ParthKohli
    • one year ago
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    \[a = \frac{dv}{dt} = v\frac{dv}{dr }\]\[a ~dr = v~ dv\]

  53. ParthKohli
    • one year ago
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    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. :|

  54. ParthKohli
    • one year ago
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    Could anyone suggest an equation to use with this?

  55. ParthKohli
    • one year ago
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    Either we could take the acceleration to be constant, or...

  56. ParthKohli
    • one year ago
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    Dan, Ganeshie... anyone?

  57. ganeshie8
    • one year ago
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    i don't know, i couldn't comprehend yet, what happens when you throw an object radially out from a satellite in orbit

  58. ganeshie8
    • one year ago
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    never thought about it before, so im gonna think about it a bit..

  59. ParthKohli
    • one year ago
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    arrey, have you ever thrown a ball vertically up? what happens? it comes back.

  60. ganeshie8
    • one year ago
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    but the satellite also got tangential velocity right

  61. ParthKohli
    • one year ago
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    then relative to the astronaut, he can throw it at a velocity that cancels out the tangential component

  62. ParthKohli
    • one year ago
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    we're mostly interested in the normal component though

  63. ganeshie8
    • one year ago
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    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 ?

  64. ParthKohli
    • one year ago
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    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!

  65. ganeshie8
    • one year ago
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    Okay I see now, so the gravity eats up the normal component of velocity completely

  66. ganeshie8
    • one year ago
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    tangential component will be same as that of Peter's

  67. ParthKohli
    • one year ago
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    of the ball? nah, it would be thrown such that it cancels out the tangential component completely.

  68. ParthKohli
    • one year ago
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    though I like dan's idea better - we can just make it have the same tangential component in the opposite direction

  69. ganeshie8
    • one year ago
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    i was thinking you both had the same idea let me scroll up read again

  70. ganeshie8
    • one year ago
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    Okay so you want to kill the tangential component of velocity and give it some normal velocity radially out, nice

  71. ParthKohli
    • one year ago
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    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

  72. ganeshie8
    • one year ago
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    Wow, I see now, that should work perfectly, at least theoretically !

  73. ganeshie8
    • one year ago
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    sorry i am just thick at learning new stuff, takes time..

  74. ParthKohli
    • one year ago
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    yes! as long as it doesn't exceed the escape velocity, we should be fine

  75. ganeshie8
    • one year ago
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    you're throwing radially out dan is throwing it with an angle so that it gets a parabolic trajectory

  76. ParthKohli
    • one year ago
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    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!

  77. ParthKohli
    • one year ago
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    get out

  78. ganeshie8
    • one year ago
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    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 \(\sqrt{2} v_{orb}\).

  79. ParthKohli
    • one year ago
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    by symmetry the first half of the motion should take 40 minutes\[a dr = v dv \]\[\int_{7\cdot 10^6 }^{r} - \frac{GM}{r^2 }dr = \int_u^0 vdv \]

  80. ParthKohli
    • one year ago
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    hmm, apparently time doesn't seem to come anywhere in the equation so far so maybe a separate one is needed for that?

  81. ParthKohli
    • one year ago
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    err why doesn't it work

  82. ParthKohli
    • one year ago
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    which equation should i use to get the time factor in

  83. ganeshie8
    • one year ago
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    can't we assume the acceleration is constant since 7000km is near earth ?

  84. ParthKohli
    • one year ago
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    we definitely can, but I'm not trying to leave anything out. it may go really far out you know...

  85. ganeshie8
    • one year ago
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    ahh 40 minutes maybe too long for flight time, we don't know yeah

  86. ParthKohli
    • one year ago
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    the flight time is 80 minutes but I divided it into two halves why is it giving me so much trouble? err

  87. ParthKohli
    • one year ago
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    |dw:1443424701627:dw|

  88. ParthKohli
    • one year ago
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    why am I not able to solve an elementary motion problem?!

  89. ParthKohli
    • one year ago
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    oh lol\[a dr = vdv\]\[-GM \int_{r_0}^r \frac{1}{r^2 } \cdot dr = \int_{u}^0 v \cdot dv\]\[2GM\left(\frac{1}{r_0} - \frac{1}{r }\right) =u^2 \]

  90. ParthKohli
    • one year ago
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    now we need to determine what \(r\) is from the time...

  91. ganeshie8
    • one year ago
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    can we use work done = change in kinetic energy power = dw/dt

  92. ParthKohli
    • one year ago
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    wow lol, the equation I just found says exactly that, decrease in kinetic energy = increase in potential energy

  93. ganeshie8
    • one year ago
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    yeah need to use that power equation somehow because that is where time comes into the picture

  94. ParthKohli
    • one year ago
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    this problem should be easier to solve than what it is right now let me ask my uncle for a moment

  95. ganeshie8
    • one year ago
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    Okay I don't have any clue how to use 40 minutes.. I'll wait :)

  96. ParthKohli
    • one year ago
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    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.

  97. ParthKohli
    • one year ago
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    @vishweshshrimali5 is also online in my friends' list. do you know him?

  98. ParthKohli
    • one year ago
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    oh my, what do you think about double integration?

  99. ganeshie8
    • one year ago
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    you want to factor in both \(r\) and \(t\) at the same time im useless here, never did these problems before

  100. ParthKohli
    • one year ago
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    Then your next project is kinematics.

  101. IrishBoy123
    • one year ago
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    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

  102. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @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*

  103. ParthKohli
    • one year ago
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    can anyone help me finish the problem now that this thing is getting the attention it deserves?

  104. Jhannybean
    • one year ago
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    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

  105. ParthKohli
    • one year ago
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    I'm going to repeat what I'm asking.|dw:1443434874247:dw|

  106. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @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

  107. ParthKohli
    • one year ago
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    Solve for \(u\) if the parameters given are:\[a(r) = -\frac{GM}{r}\]\[t = 80\cdot 60 ~\rm sec\]

  108. ParthKohli
    • one year ago
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    @imqwerty ^^^

  109. thomas5267
    • one year ago
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    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.

  110. thomas5267
    • one year ago
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    The chances are you can throw it with any velocity as long as you aim in the right direction.

  111. ganeshie8
    • one year ago
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    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...

  112. imqwerty
    • one year ago
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    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

  113. thomas5267
    • one year ago
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    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?

  114. ganeshie8
    • one year ago
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    I think our goal here is to find \(v(t_0)\), why do we require it to be 0 ?

  115. thomas5267
    • one year ago
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    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.

  116. thomas5267
    • one year ago
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    Lagrangian mechanics? \[ L=T-V\quad \text{T is total kinetic energy, V is total potential.}\\ L=\frac{1}{2}mv^2+\frac{Gm}{r}=0\text{ by energy conservation.}\\ \frac{\partial L}{\partial \mathbf{r}} - \frac{d}{dt}\frac{\partial L}{\partial \dot{\mathbf{r}}} =0\\ r=\|\mathbf{r}(t)\|\\ v^2=\mathbf{v}(t)\cdot \mathbf{v}(t)=\mathbf{\dot{r}}(t)\cdot\mathbf{\dot{r}}(t) \]

  117. ganeshie8
    • one year ago
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    @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|

  118. ganeshie8
    • one year ago
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    Interesting, whats the trajectory of lunchbox that you're assuming @thomas5267

  119. thomas5267
    • one year ago
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    That is the equation that yields the trajectory if I am not mistaken.

  120. thomas5267
    • one year ago
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    The partial derivatives with respect to the vectors are componentwise partial derivatives. https://en.m.wikipedia.org/wiki/Lagrangian_mechanics#Equations_of_motion

  121. IrishBoy123
    • one year ago
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    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.

  122. imqwerty
    • one year ago
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    yes @ganeshie8 and @thomas are u throwing the box like this-|dw:1443438908909:dw|