Calculating v in 0.5mv^2: Right or Wrong?

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Homework Help Overview

The discussion revolves around calculating the escape velocity of a rocket in relation to the gravitational influences of the Moon and Jupiter. The original poster presents a formula involving gravitational constants and distances, aiming to reconcile their calculated value with a reference from a textbook.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of treating celestial bodies as point masses and question the distances used in calculations. There are attempts to clarify the setup regarding the position of the rocket relative to the Moon and Jupiter, as well as discussions on the accuracy of the provided values.

Discussion Status

Multiple interpretations of the problem are being explored, with participants providing varying calculations and questioning the assumptions made. Some participants suggest re-evaluating the distances and the treatment of the celestial bodies, while others express uncertainty about the correctness of their calculations.

Contextual Notes

There is mention of potential misprints in the problem's provided values, and participants are encouraged to verify their calculations step-by-step. The discussion reflects a mix of confidence and confusion regarding the expected results.

UrbanXrisis
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The question is http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

[tex]0.5mv^2=\frac{GmM_{moon}}{R_{moon}}+\frac{GmM_{jupiter}}{R_{jupiter}+x}[/tex]
x=distance between Jupiter and the moon
[tex]v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}+x}}[/tex]

[tex]v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{6.99E7+1.071E9}}[/tex]

[tex]v=17.657km/s[/tex]

when I subbed in the numbers, I get about 17km/s. The books gives 15km/s, am I doing this right?
 
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I think you should consider the planets as point masses, particularly Jupiter on the right side of your first line. I would think that the distance 'x' is from center to center.
 
that would make the velocity bigger, not smaller, I already tired that but got +18km/s
 
Take into account that the rocket is on the opposite side of the moon from jupiter. so the distance between Jupiter and the rocket would be the distance between them plus a radius of the moon
 
UrbanXrisis said:
that would make the velocity bigger, not smaller, I already tired that but got +18km/s
Nevertheless, this is the correct answer (~18 km/sec).

You can neglect the radius of Ganymede (it's 3 orders of magnitude smaller) when calculating the distance from Jupe's center.
 
whozum said:
Take into account that the rocket is on the opposite side of the moon from jupiter. so the distance between Jupiter and the rocket would be the distance between them plus a radius of the moon

I already tried that too.. the radius of the moon is too small to decrease the escape velocity by 2km/s

Gokul43201 said:
Nevertheless, this is the correct answer (~18 km/sec).

You can neglect the radius of Ganymede (it's 3 orders of magnitude smaller) when calculating the distance from Jupe's center.

my book tells me an answer of 15.6 km/s, unless you're saying that it was a missprint?
 
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[tex]v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}}}[/tex]

[tex]v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}[/tex]

[tex]v=18136m/s[/tex]
 
Last edited:
UrbanXrisis said:
my book tells me an answer of 15.6 km/s, unless you're saying that it was a missprint?
Either that, or perhaps, one of the numbers provided in the problem is a misprint.
 
so you would agree that the escape velocity is indeed 18.136km/s?
 
  • #10
No, I just redid the calculation and got 15.6 km/sec. Recheck your calculation - step by step. Post numbers here if necessary.
 
  • #11
Distance = Radius of Jupiter + Distance between 2 planets + 2* Radius of Moon.
Viet Dao,
 
  • #12
VietDao29 said:
Distance = Radius of Jupiter + Distance between 2 planets + 2* Radius of Moon.
Viet Dao,
I don't think so. The specified distance between Jupe and Gany is the distance between their centers.
 
  • #13
UrbanXrisis said:
[tex]v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}}}[/tex]

[tex]v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}[/tex]

what this the right path? I've been punching numbers into my calc for a while now, I don't see how you got 15.6 km/s
 
  • #14
You're punching numbers wrong, if you have maple, you can do the following:

>v=sqrt((2*G*M)/R+(2*G*N/P));

> subs(G=6.67*10^(-11),M=1.495*10^23,N=1.9*10^27,R=2.64*10^6,P=1.071*10^9,%);
 
  • #15
[tex]v=\sqrt{\frac{2GM_{moon}}{R_{moon}}+\frac{2GM_{jupiter}}{R_{jupiter}}}[/tex]

[tex]v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}+\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}[/tex]

[tex]v=\sqrt{\frac{2*6.673x10^{-11}*1.495E23kg}{2.64E6m}}+\sqrt{\frac{2*6.673x10^{-11}*1.9E27kg}{1.071E9}}[/tex]

[tex]v=\sqrt{\frac{1.995227*10^{13}}{2.64x10^6}}+\sqrt{\frac{2.5357*10^{17}}{1.071x10^9}}[/tex]

[tex]v=2.749x10^3 + 1.5387x10^4 = 1.8156x10^4[/tex]

still the same answers :confused:
 
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  • #16
[tex]\sqrt{a + b} = \sqrt{a} + \sprt{b}[/tex]
Are you sure?
Viet Dao,
 
  • #17
I see my mistake... thanks

that took me way too long
 
  • #18
You can't split up the square root.
 
  • #19
yeah... what a stupid mistake... arggg
 
  • #20
dont worry, i once embarassed myself in a class by telling the teacher to split up 1/(x^2+x) into 1/x^2 + 1/x.

Atleast your mistake was infront of 3 and not 40.
 

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