GRE Problem #66: orbit of the spacecraft about the Sun?

[quantumworld]

Dear reader,
here is a neat problem, but kindof over the edge.

when it is about the same distance from the Sun as is Jupiter, a spacecraft on a mission to the outer planets has a speed that is 1.5 times the speed of Jupiter in its orbit. Which of the following describes the orbit of the spacecraft about the Sun?
(A) Spiral
(B) Circle
(C) Ellipse
(D) Parabola
(E) Hyperbola

The way I tried to tackle this problem, is by saying that the eccentricity is more than one (it is 1.5), thus it is a hyperbola, which is the correct answer, BUT I don't think it is enough to know that, I mean my answer assumes that Jupiter's orbit is a hyperbola (eccentricity = 1), but I am not sure if that is correct, or even more, I don't know why do I need to know the orbit's of planets in general . Your contribution is very valuable.

Many thanks.

 

[Tide]

No, the problem (or answer) does not assume that the orbit of Jupiter is hyperbolic.

Think escape velocity! :-)

 

[HallsofIvy]

1. The orbit of Jupiter definitely not hyperbolic!

2. Eccentricity 1 is a parabola, not a hyperbola.
(But the orbit of Jupiter is also not a parabola, it is an ellipse, pretty close to a circle- its eccentricity is very close to 0, not 1.)

 

[TenaliRaman]

Hmm thinking over it a bit, i think quantum is right tho i would be more tended towards thinking that orbit of Jupiter is elliptical.

what can we say abt orbit shapes in the foll cases?
case1 > what if the
velocity of spacecraft = velocity required to maintain circular motion
Answer is straightforward

case2 > what if the
velocity of spacecraft > velocity required to maintain circular motion but
velocity of spacecraft < escape velocity
Answer is straightforward

case3 > what if the
velocity of spacecraft = escape velocity
this is tricky , we need to think in terms of geometry

case 4 > what if the
velocity of spacecraft > escape velocity
again tricky ,need to think in terms of geometry.

Apparently our assumption of the orbit of Jupiter around sun does affect the choice we make. Even elliptical orbit assumption could result in answer of case3 or case4 because we don't know whether 1.5 times Jupiter velocity is going to push it beyond escape velocity.

P.S : Unless ofcourse if we are provided with mass of Jupiter and orbit radius and the actual Jupiter velocity

 

[quantumworld]

No, the problem (or answer) does not assume that the orbit of Jupiter is hyperbolic.

Think escape velocity! :-)

Tide,
Could you please give me little more hints...

 

[quantumworld]

1. The orbit of Jupiter definitely not hyperbolic!

2. Eccentricity 1 is a parabola, not a hyperbola.
(But the orbit of Jupiter is also not a parabola, it is an ellipse, pretty close to a circle- its eccentricity is very close to 0, not 1.)

HallsofIvy,
it was a typo saying that e=1 is a hyperbola, I meant a prabola, sorry about that
thanks though, I wasn't sure about jupiter's orbit, but I guess that because we still see jupiter, it must be in a circle or an ellipse, otherwise, we won't be able to see it, please correct me if I am wrong...

 

[quantumworld]

Tenaliraman,
, I do agree with what u said, and I am still confused

 

[Tide]

Tide,
Could you please give me little more hints...

Jupiter's orbit is nearly circular. If it were perfectly circular than the escape velocity of the spacecraft (with respect to the SUN!) would be \sqrt 2 times Jupiter's speed. Since 1.5 times Jupiter's speed is substantially greater than the escape velocity the speed of the craft will be greater than the escape velocity even allowing for some slight eccentricity in Jupiter's orbit. I don't believe it's called for in this problem but you could, if you wanted to, verify using the actual eccentricity of Jupiter's orbit.

The spacecraft is moving faster than the escape velocity so its "orbit" about the sun must be hyperbolic!

 

[quantumworld]

Thanks Tide,

I found this online, so I thought of posting it, in order to clarify the difference between parabolic and hyperbolic orbit as related to the escape velocity...

If the orbit is perfectly circular, the magnitude of the velocity is constant and given by

Vorb = sqrt(GM/r),
where G is the gravitational constant, M is the mass of the gravitating body, and r is the radius of the orbit. An object moving faster than circular velocity will enter an elliptical orbit with a velocity at any point determined by Kepler's laws of planetary motion. If the object moves faster still, it will travel at escape velocity along a parabolic orbit or beyond escape velocity in a hyperbolic orbit.