What is Pluto's speed at the most distant point in its orbit?

[JJBladester]

Homework Statement

Pluto moves in a fairly elliptical orbit around the sun. Pluto's speed at its closest approach of 4.43x10⁹km is 6.12km/s.

What is Pluto's speed at the most distant point in its orbit, where it is 7.30x10⁹km from the sun?

Homework Equations

Conservation of energy:
$$K_{2} + U_{2} = K_{1} + U_{1}$$

The Attempt at a Solution

$$M_{p}$$ = Mass of Pluto (actually cancels out when re-arranging the eq to get $$v_{2}$$)

$$M_{s}$$ = Mass of Sun

$$\frac{1}{2}M_{p}v_{2}^{2} - \frac{GM_{s}M_{p}}{r_{2}} = \frac{1}{2}M_{p}v_{1}^{2} - \frac{GM_{s}M_{p}}{r_{1}}$$

After fiddling with the equation above, I get:

$$v_{2} = v_{1} + \sqrt{\frac{2GM_{s}}{r_{2}-r_{1}}}$$

The correct answer is 3.71 km/s, but my answer comes out differently. I'm using the following numbers in the equation above:

$$v_{2} = 6.12\cdot10^{3}m/s + \sqrt{\frac{2(6.67\cdot10^{-11})(1.99\cdot10^{30})}{7.30\cdot10^{12} - 4.43\cdot10^{12}}} = 15.7km/s$$

Perhaps a 2nd set of eyes could find where I went wrong on this.

 

[tiny-tim]

Hi JJBladester!

$$\frac{1}{2}M_{p}v_{2}^{2} - \frac{GM_{s}M_{p}}{r_{2}} = \frac{1}{2}M_{p}v_{1}^{2} - \frac{GM_{s}M_{p}}{r_{1}}$$

After fiddling with the equation above, I get:

$$v_{2} = v_{1} + \sqrt{\frac{2GM_{s}}{r_{2}-r_{1}}}$$

I'm sorry, but this is wrong in two ways …

1/r₁ - 1/r₂ is not 1/(r₁ - r₂)

and v₁ - v₂ is not √(v₁² - v₂²)

 

[D H]

Conservation of energy will work, but conservation of angular momentum will be much easier for this problem.

Note that in general, conservation of angular momentum can only tell you about the component of velocity normal to the radial vector. However, at perihelion and apohelion, the velocity vector is normal to the radial vector.

 

[tiny-tim]

Conservation of energy will work, but conservation of angular momentum will be much easier for this problem.

ooh, i didn't think of that!

 

[JJBladester]

Thanks all... Got it figured out now. I guess trying to do physics too late at night is a bad idea... Too many easy math errors can be made. Thanks for the good replies. That's why I love this forum... because the help is just as good as if I were taking an on-campus course.

 

[JJBladester]

Conservation of energy will work, but conservation of angular momentum will be much easier for this problem.

Note that in general, conservation of angular momentum can only tell you about the component of velocity normal to the radial vector. However, at perihelion and apohelion, the velocity vector is normal to the radial vector.

Wow... Great observation! The problem took under 5 minutes to solve this way and I didn't have to spend a bunch of time converting km's to m's due to the G constant being in a different set of units. Thanks again.