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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- ganeshie8

|dw:1443419836472:dw|

- ganeshie8

\(7000km\) is the distance from the "center of earth" to Mary/Peter's orbit

- dan815

|dw:1443419927131:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ganeshie8

they are orbiting the earth, so angular velocity \(\omega\) cannot be \(0\), right ?

- dan815

change in w

- dan815

what im asking is are they orbitting at same angular velocity

- ganeshie8

Oh ok, let me add that info, sorry... i must give the distance between Peter and Mary

- ganeshie8

I'll update that in the main question quick..

- 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

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

if we apply normal physics then this is really just a parabollic equation problem

- dan815

|dw:1443420214293:dw|

- ganeshie8

|dw:1443420236862:dw|

- ParthKohli

We're probably gonna have to use the conservation of angular momentum here. Probably.

- dan815

|dw:1443420284948:dw|

- ParthKohli

So it's not necessary that the lunch-box follows a circular trajectory - that right?

- 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

that is if dw/dt=0 meaning their distance apart is constant

- ganeshie8

lunch box follows whatever the trajectory that gravity decides, for sure, its not a parabola or circle

- dan815

okay so just answer that 1 questio nis the distance between them constant

- ParthKohli

Yes, but it depends on the angle with which the thing is thrown, right?

- 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

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

okay so ya like parth is suggesting, there are multiple solutions based on the angle chosen

- ganeshie8

there is only one solution
depends on what trajectory an object in orbit takes when you increase/decrease its velocity

- 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

|dw:1443420881110:dw|

- ganeshie8

Eactly!
|dw:1443420863876:dw|

- ParthKohli

not tangentially but normally

- 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

wait, are you talking to me?

- ganeshie8

talking to this one :
|dw:1443421091159:dw|

- 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

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

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

okay heres a solution that must work then i guess

- dan815

how about throwing 2*Velocity backwards

- ParthKohli

yes, but that's horizontal. we're throwing it vertically away from the earth.

- ParthKohli

that's true. 2*velocity relative to the astronaut. it follows a clockwise orbit in that case. nice.

- dan815

|dw:1443421483568:dw|

- ParthKohli

|dw:1443421559981:dw|

- 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

I think, it can never come back to the same orbit because the "speed" of an object uniquely determines the orbit radius

- dan815

if u ask me its better to ignore this physics and just write out pure equations

- dan815

we might fight all kinds of solutions just working with these differential equations

- dan815

start with the path equation of mary
we can use peter just for initial conditions

- ganeshie8

Alright, for simplicity we may assume the period of both peter and mary is 100 minutes.

- ParthKohli

|dw:1443421917404:dw|

- 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

|dw:1443422101849:dw|

- ganeshie8

right, that should come around 100 minutes

- ParthKohli

\[a = \frac{dv}{dt} = v\frac{dv}{dr }\]\[a ~dr = v~ dv\]

- 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

Could anyone suggest an equation to use with this?

- ParthKohli

Either we could take the acceleration to be constant, or...

- ParthKohli

Dan, Ganeshie... anyone?

- 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

never thought about it before, so im gonna think about it a bit..

- ParthKohli

arrey, have you ever thrown a ball vertically up? what happens? it comes back.

- ganeshie8

but the satellite also got tangential velocity right

- ParthKohli

then relative to the astronaut, he can throw it at a velocity that cancels out the tangential component

- ParthKohli

we're mostly interested in the normal component though

- 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

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

Okay I see now, so the gravity eats up the normal component of velocity completely

- ganeshie8

tangential component will be same as that of Peter's

- ParthKohli

of the ball? nah, it would be thrown such that it cancels out the tangential component completely.

- ParthKohli

though I like dan's idea better - we can just make it have the same tangential component in the opposite direction

- ganeshie8

i was thinking you both had the same idea
let me scroll up read again

- ganeshie8

Okay so you want to kill the tangential component of velocity and give it some normal velocity radially out, nice

- 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

Wow, I see now, that should work perfectly, at least theoretically !

- ganeshie8

sorry i am just thick at learning new stuff, takes time..

- ParthKohli

yes! as long as it doesn't exceed the escape velocity, we should be fine

- ganeshie8

you're throwing radially out
dan is throwing it with an angle so that it gets a parabolic trajectory

- 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

get out

- 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

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

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

- ParthKohli

err why doesn't it work

- ParthKohli

which equation should i use to get the time factor in

- ganeshie8

can't we assume the acceleration is constant since 7000km is near earth ?

- ParthKohli

we definitely can, but I'm not trying to leave anything out. it may go really far out you know...

- ganeshie8

ahh 40 minutes maybe too long for flight time, we don't know yeah

- ParthKohli

the flight time is 80 minutes but I divided it into two halves
why is it giving me so much trouble? err

- ParthKohli

|dw:1443424701627:dw|

- ParthKohli

why am I not able to solve an elementary motion problem?!

- 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

now we need to determine what \(r\) is from the time...

- ganeshie8

can we use
work done = change in kinetic energy
power = dw/dt

- ParthKohli

wow lol, the equation I just found says exactly that,
decrease in kinetic energy = increase in potential energy

- ganeshie8

yeah need to use that power equation somehow
because that is where time comes into the picture

- ParthKohli

this problem should be easier to solve than what it is right now
let me ask my uncle for a moment

- ganeshie8

Okay I don't have any clue how to use 40 minutes.. I'll wait :)

- 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

@vishweshshrimali5 is also online in my friends' list. do you know him?

- ParthKohli

oh my, what do you think about double integration?

- ganeshie8

you want to factor in both \(r\) and \(t\) at the same time
im useless here, never did these problems before

- ParthKohli

Then your next project is kinematics.

- 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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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

can anyone help me finish the problem now that this thing is getting the attention it deserves?

- 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

I'm going to repeat what I'm asking.|dw:1443434874247:dw|

- 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

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

- ParthKohli

@imqwerty ^^^

- 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

The chances are you can throw it with any velocity as long as you aim in the right direction.

- 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

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

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

I think our goal here is to find \(v(t_0)\),
why do we require it to be 0 ?

- 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

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

@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

Interesting, whats the trajectory of lunchbox that you're assuming @thomas5267

- thomas5267

That is the equation that yields the trajectory if I am not mistaken.

- 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

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

yes @ganeshie8 and @thomas are u throwing the box like this-|dw:1443438908909:dw|

- 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

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

I mean \(\dfrac{d^2r}{dt^2}\) not \(\dfrac{d^2r}{dt}\).

- 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

\[
\frac{1}{2}t^2-\frac{2}{5}Pt-\frac{7000000^4}{12GM}=-\frac{1}{12}r(t)^4\\
-6t^2+\frac{24}{5}Pt+\frac{7000000^4}{GM}=r(t)^4
\]
Take the quartic root?

- thomas5267

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

Can't paste the url. Type Integrate[1/Sqrt[2GM/r+C],r] into Wolfram Alpha.

- 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

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!

Looking for something else?

Not the answer you are looking for? Search for more explanations.