GR9677 #66



Alternate Solutions 
The_Duck 20100705 15:22:56  Others have posted this in a more complicated way, but here's a very simple way to think about it:
Take Jupiter's orbit to be circular. Then the virial theorem says that the ratio of the magnitude of its kinetic and (negative) potential energy is 1:2. The spacecraft is going at 1.5 times Jupiter's speed so its ratio is 1.5^2:2 = 2.25:2. Since it has more kinetic than (negative) potential energy is not bound: it will escape the solar system and is thus in a hyperbolic orbit.
If you remember all of this, the trickiest part is remembering to square 1.5. The virial theorem is very nice to know here; it lets you avoid mucking around with any sort of explicit potential, real or effective.   GREview 20090918 00:14:03  The spacecraft is "on a mission to the outer planets" so why would be in an orbit that doesn't allow it to escape? Thus it's either D or E. Given that D is a very special case, one chooses E.   apps 20090808 05:57:11  I believe it could be solved in a simpler way.
We know that Jupiter orbits around the Sun and doesn't fall. So its speed is at least V_c=$\sqrt{GM/R}.
One should have kinetic energy equals or greater than potential energy of gravitational field in order to leave the Sun:
mV^2/2 >= GmM/R
V >= $\sqrt{2*GM/R} >= $\sqrt{2} * V_c ("=" means parabola)
1.5 > $\sqrt{2} means that spacecraft has sufficient energy to leave   OrbitGirl 20071002 10:39:35  It doesn't have to be that complicated.
One could just recognize that an object traveling at 1.5 times the speed of Jupiter in the same vicinity would be traveling in excess of escape velocity with respect to the Sun. Speed greater than escape velocity results in a hyperbolic orbit. QED.
You can check with NASA if you don't believe me: http://www.hq.nasa.gov/office/pao/History/conghand/traject.htm
Blue Quark 20071027 17:24:23 
The trouble is finding the escape velocity. The easiest way is to recognize that the acceleration needed for circular orbits is centripetal acceleration equal to a=/r
Then ma=G(M*m)/ m/r=G(M*m)/
Solve and find =
To find escape velocity, set kinetic energy equal to gravitational potential energy.
1/2*m*=G*(M*m)/r
Solve and find =
The end result is that = *
Since 1.5> the velocity is greater than the escape velocity and hence it travels a hyperbolic path about the sun.

  sharpstones 20070405 07:29:46  Ok, here's how to do the problem. Along with memorizing the Effective Potential you really should memorize the total energy of a body in a circular orbit:
And the Virial Theorem:
And the classification of orbits
=> circular orbit
=> elliptical orbits
=> parabolic orbit
$E > 0 $=> hyperbolic orbit
On to the problem: Since Jupiter has an elliptical orbit we know that:
=>
Now for the spaceship you know that
At this point the approximation you make is that Jupiter's orbit is not very far from a circle so E is close to E_c and that means that
is close to
so for sure multiplying by 2.25 will make it greater then which will bring you to the hyperbolic regime  

Comments 
aziza 20140811 14:32:33  I thought this was a Hohmann orbit transfer question..so I chose (c) ellipse :(   phoxdie 20101110 16:36:46  Very curious. In the "offical" solution it is stated that the total energy of the spacecraft is:
and for Jupiter it is:
then the claim is made that the spacecraft energy is greater than Jupiters based solely on the fact that it has a greater velocity. How can this be correct considering that the spacecraft must have a mass much much much less than that of Jupiter? Rereading what you have I noticed that the so in this case I believe the correct reasoning for the ship having a greater energy than Jupiter is because it has a much smaller mass. Since both the planet and the ship are at roughly the same distance from the Sun any dependence can be neglected.
I don't disagree with the answer, only the way in which it is explained.   The_Duck 20100705 15:22:56  Others have posted this in a more complicated way, but here's a very simple way to think about it:
Take Jupiter's orbit to be circular. Then the virial theorem says that the ratio of the magnitude of its kinetic and (negative) potential energy is 1:2. The spacecraft is going at 1.5 times Jupiter's speed so its ratio is 1.5^2:2 = 2.25:2. Since it has more kinetic than (negative) potential energy is not bound: it will escape the solar system and is thus in a hyperbolic orbit.
If you remember all of this, the trickiest part is remembering to square 1.5. The virial theorem is very nice to know here; it lets you avoid mucking around with any sort of explicit potential, real or effective.
Quark 20111019 14:47:25 
I think this solution is the quickest and easiest to understand conceptually. The only set back would be recognizing when you can use the virial theorem.

Quark 20111019 15:30:32 
Just to make things a bit more explicit for those that are not familiar with the virial theorem. The virial theorem gives a relation between the total kinetic energy of a system to the total potential energy of that system. Mathematically,
2=n
where n comes from V(r)=a. Since the force here is just the gravitational force which is proportional to , we have that the potential is proportional to . Therefore our ratio of the kinetic energy to the potential energy for Jupiter is
2=
where the rightside is positive since the potential is negative and n=1. Now if we solve for the total kinetic energy we get that
which states that the Kinetic energy is half of the potential energy and therefore Jupiter doesn't have enough kinetic energy to escape the sun's gravitational pull (hence it's orbit around the sun). Thus the ratio of the potential energy to the kinetic energy of Jupiter is 2:1 Now we want to find this ratio for the spaceship. Since the spacecraft is traveling at 1.5 times the speed of Jupiter, its will be larger than that of Jupiter's since kinetic energy is proportional to . Our new ratio of potential energy to kinetic energy is 2:1(2.25) showing that the spacecraft has more kinetic energy than potential energy and will thus escape the sun.

dragore 20120626 15:38:40 
^ You seem to forget that also depends on the mass. It would be nonsense to conclude that the kinetic energy of a spaceship could be larger than that of Jupiter.

  GREview 20090918 00:14:03  The spacecraft is "on a mission to the outer planets" so why would be in an orbit that doesn't allow it to escape? Thus it's either D or E. Given that D is a very special case, one chooses E.
someone 20111106 16:32:13 
I like this solution.

Nebula 20150916 22:12:36 
Easily the quickest solution.

  apps 20090808 05:57:11  I believe it could be solved in a simpler way.
We know that Jupiter orbits around the Sun and doesn't fall. So its speed is at least V_c=$\sqrt{GM/R}.
One should have kinetic energy equals or greater than potential energy of gravitational field in order to leave the Sun:
mV^2/2 >= GmM/R
V >= $\sqrt{2*GM/R} >= $\sqrt{2} * V_c ("=" means parabola)
1.5 > $\sqrt{2} means that spacecraft has sufficient energy to leave
archard 20100930 19:56:27 
This is what ETS is getting at I think, and probably the best way of going about getting the right answer.

  OrbitGirl 20071002 10:39:35  It doesn't have to be that complicated.
One could just recognize that an object traveling at 1.5 times the speed of Jupiter in the same vicinity would be traveling in excess of escape velocity with respect to the Sun. Speed greater than escape velocity results in a hyperbolic orbit. QED.
You can check with NASA if you don't believe me: http://www.hq.nasa.gov/office/pao/History/conghand/traject.htm
Blue Quark 20071027 17:24:23 
The trouble is finding the escape velocity. The easiest way is to recognize that the acceleration needed for circular orbits is centripetal acceleration equal to a=/r
Then ma=G(M*m)/ m/r=G(M*m)/
Solve and find =
To find escape velocity, set kinetic energy equal to gravitational potential energy.
1/2*m*=G*(M*m)/r
Solve and find =
The end result is that = *
Since 1.5> the velocity is greater than the escape velocity and hence it travels a hyperbolic path about the sun.

gliese876d 20081011 21:04:37 
blue quark, that's a great way of looking at it. also, it just sort of seems logical, doesn't it, that if the spacecraft is bound for the outer planets, it probably needs a hyperbolic orbit

motek 20081021 09:45:42 
Blue's solution is better because it does not assume that the mass of the spacecraft is equal to the mass of Jupiter, which is wring and is assumed by the first solution

hanin 20090929 22:10:40 
@Blue Quark
Nice solution!
Small question: what if the speed is between circular and escape velocity? An ellipse orbit? > that way the number 1.5 in the problem is essential

shak 20100816 21:09:25 
good analysis!!!!!thank u

flyboy621 20101102 21:53:55 
Good solution, Blue Quark. Hanin, I think your comment highlights why you have to do the calculation. If the spaceship's speed had been 1.3 times Jupiter's (less than that is), it would be an elliptical orbit.

  sharpstones 20070405 07:29:46  Ok, here's how to do the problem. Along with memorizing the Effective Potential you really should memorize the total energy of a body in a circular orbit:
And the Virial Theorem:
And the classification of orbits
=> circular orbit
=> elliptical orbits
=> parabolic orbit
$E > 0 $=> hyperbolic orbit
On to the problem: Since Jupiter has an elliptical orbit we know that:
=>
Now for the spaceship you know that
At this point the approximation you make is that Jupiter's orbit is not very far from a circle so E is close to E_c and that means that
is close to
so for sure multiplying by 2.25 will make it greater then which will bring you to the hyperbolic regime   sharpstones 20070405 06:53:51  I also don't understand this reasoning considering in the above equations for the KE the same mass is used for Jupiter and the spaceship??   cyberdeathreaper 20070128 11:23:56  I think I'm not understanding something here...
In part (B) you say "the energy of Jupiter is greater than that of the spaceship"
But in part (C) you state that while the energy of Jupiter falls between Vmin and 0, the total energy of the spacecraft must be greater than 0, implying the spacecraft has more energy than Jupiter.
Both can't be true  intuition tells me Jupiter has more energy, but I'm unsure how that leads to the correct answer.
tau1777 20081107 11:56:00 
so remembering
Energy, Orbit
E > 0, hyperbola
E = 0 , parabola
Vmin < E<0 ellipse
E=Vmin, Circle
i was assuming that jupiter has ellipitical orbit, then you know its either parabola or hyperbola. but moving so fast that KE term is large to most likely hyperbola. this solution is probably unsettling as is requires a bit of luck. but once you can get it down to two choice, it pays off to take a shot.

  quarky 20051209 17:19:33  How can you know the energy of the spaceship without knowing its mass?
rawr 20090925 19:51:01 
You know that it's A LOT less massive than jupiter. that's all you need for a basic analysis

 

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