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Mechanics}Effective Potential


One can solve this problem by remembering the effective potential curve V_{eff}(r) = V(r)+L^2/(2mr^2). For the gravitational potential, one has V(r) \propto - 1/r.

The total energy of the spaceship is E_s = 1/2 m(1.5v_J)^2+V_{eff}, while the total energy of Jupiter is E_J = 1/2 m(v_J)^2 + V_{eff}

(A) A spiral orbit occurs when E<V_{min}, which corresponds to v_s<<v_J.

(B) A circular orbit occurs only when E = V_{min}. Since the energy of Jupiter is greater than that of the spaceship--and (see below) since Jupiter itself has E>V_{min}, the spaceship must have E>V_{min}.

(C) An ellipse occurs for V_{min}<E<0. Planets orbit in ellipses. However, since the speed of the ship is greater than Jupiter's orbit speed by a good bit, one assumes its total energy is E>0.

(D) A parabolic orbit occurs for E=0. The condition is much too stringent.

(E) A hyperbolic orbit occurs for E>0. See (C). Since E_s>0>E_J, this is it.


See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
GREview
2009-09-18 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.Alternate Solution - Unverified
apps
2009-08-08 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		Alternate Solution - Unverified
OrbitGirl
2007-10-02 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
2007-10-27 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=v^2/r
Then ma=G(M*m)/r^2 \Rightarrow mv^2/r=G(M*m)/r^2
Solve and find v_{circular}=\sqrt{GM/r}

To find escape velocity, set kinetic energy equal to gravitational potential energy.
1/2*m*v^2_{escape}=G*(M*m)/r
Solve and find v_{escape}=\sqrt{2GM/r}

The end result is that v_{escape} = \sqrt{2}*v_{circular}
Since 1.5>\sqrt{2} the velocity is greater than the escape velocity and hence it travels a hyperbolic path about the sun.
Alternate Solution - Unverified
sharpstones
2007-04-05 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:

E_c = - \frac{GMm}{2R}

And the Virial Theorem:
\lt KE\g = - \lt E \g
\lt U \g = 2 \lt E \g

And the classification of orbits
E = E_c => circular orbit
E_c \lt E \lt = => elliptical orbits
E = 0 => parabolic orbit
$E > 0 $=> hyperbolic orbit

On to the problem: Since Jupiter has an elliptical orbit we know that:

\lt KE\g = \frac{1}{2}m_j{v_j}^2 \ \lt \ \frac{GMm_j}{2R}

=>
\frac{1}{2}{v_j}^2 \ \lt \ \frac{GM}{2R}

Now for the spaceship you know that
\frac{1}{2}{v_s}^2 = \frac{1}{2}{v_j}^2 * \frac{9}{4}\frac{1}{2}{v_j}^2 * 2.25

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

\frac{1}{2}{v_j}^2 is close to \frac{GM}{2R}

so for sure multiplying \frac{1}{2}{v_j}^2 by 2.25 will make it greater then \frac{GM}{2R} which will bring you to the hyperbolic regime		Alternate Solution - Unverified
Comments
GREview
2009-09-18 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.Alternate Solution - Unverified
apps
2009-08-08 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		Alternate Solution - Unverified
OrbitGirl
2007-10-02 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
2007-10-27 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=v^2/r
Then ma=G(M*m)/r^2 \Rightarrow mv^2/r=G(M*m)/r^2
Solve and find v_{circular}=\sqrt{GM/r}

To find escape velocity, set kinetic energy equal to gravitational potential energy.
1/2*m*v^2_{escape}=G*(M*m)/r
Solve and find v_{escape}=\sqrt{2GM/r}

The end result is that v_{escape} = \sqrt{2}*v_{circular}
Since 1.5>\sqrt{2} the velocity is greater than the escape velocity and hence it travels a hyperbolic path about the sun.
gliese876d
2008-10-11 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
2008-10-21 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
2009-09-29 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
Alternate Solution - Unverified
sharpstones
2007-04-05 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:

E_c = - \frac{GMm}{2R}

And the Virial Theorem:
\lt KE\g = - \lt E \g
\lt U \g = 2 \lt E \g

And the classification of orbits
E = E_c => circular orbit
E_c \lt E \lt = => elliptical orbits
E = 0 => parabolic orbit
$E > 0 $=> hyperbolic orbit

On to the problem: Since Jupiter has an elliptical orbit we know that:

\lt KE\g = \frac{1}{2}m_j{v_j}^2 \ \lt \ \frac{GMm_j}{2R}

=>
\frac{1}{2}{v_j}^2 \ \lt \ \frac{GM}{2R}

Now for the spaceship you know that
\frac{1}{2}{v_s}^2 = \frac{1}{2}{v_j}^2 * \frac{9}{4}\frac{1}{2}{v_j}^2 * 2.25

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

\frac{1}{2}{v_j}^2 is close to \frac{GM}{2R}

so for sure multiplying \frac{1}{2}{v_j}^2 by 2.25 will make it greater then \frac{GM}{2R} which will bring you to the hyperbolic regime		Alternate Solution - Unverified
sharpstones
2007-04-05 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??NEC
cyberdeathreaper
2007-01-28 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
2008-11-07 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 un-settling as is requires a bit of luck. but once you can get it down to two choice, it pays off to take a shot.
Answered Question!
quarky
2005-12-09 17:19:33		How can you know the energy of the spaceship without knowing its mass?
rawr
2009-09-25 19:51:01		 You know that it's A LOT less massive than jupiter. that's all you need for a basic analysis
NEC

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