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

  • Thread starter UrbanXrisis
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In summary, the question is asking if the escape velocity for a rocket is 18.136 km/s. The answer is yes, the escape velocity is 18.136 km/s.
  • #1
UrbanXrisis
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1
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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  • #2
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.
 
  • #3
that would make the velocity bigger, not smaller, I already tired that but got +18km/s
 
  • #4
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
 
  • #5
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.
 
  • #6
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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  • #7
[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]
 
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  • #8
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.
 
  • #9
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.
 

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

1. What does the equation 0.5mv^2 represent in physics?

The equation 0.5mv^2 represents the kinetic energy of an object, where m is the mass of the object and v is its velocity.

2. Is the equation 0.5mv^2 always applicable?

No, the equation 0.5mv^2 is only applicable for objects that have a constant mass and are moving at a constant velocity.

3. Can the equation 0.5mv^2 be used for objects with varying mass or velocity?

No, the equation 0.5mv^2 only applies to objects with a constant mass and velocity. For objects with varying mass or velocity, the equation needs to be modified or a different equation should be used.

4. Is the equation 0.5mv^2 only applicable for linear motion?

No, the equation 0.5mv^2 can also be used for rotational motion, as long as the object's mass and velocity remain constant.

5. Can the equation 0.5mv^2 be used to calculate potential energy?

No, the equation 0.5mv^2 only calculates the kinetic energy of an object. To calculate potential energy, a different equation such as mgh (mass x acceleration due to gravity x height) should be used.

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