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Escape velocity |
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| Jan2-05, 12:08 PM | #1 |
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Escape velocity
I can't seem to obtain the right answer using the escape velocity formula. I think my problem is with the gravitational constant. I would like to find the escape velocity of the sun.
[tex]v_e=\sqrt{\frac{2GM}{r}}[/tex] or [tex]v_e=\sqrt{\frac{2\mu}{r}}[/tex] [tex]v_e_\odot=617.54(km/s)[/tex] [tex]v_e_\odot=\sqrt{\frac{2*6.673e-11*1.9891e30(kg)}{696000(km)}}[/tex] [tex]v_e_\odot=\sqrt{~3.8141e14}\neq 617.54(km/s)[/tex] What am I doing wrong? Thanks |
| Jan2-05, 12:39 PM | #2 |
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Okay,i'll make some approximations,but i'll get the result and the order of magnitude.
[tex]v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}\sim \sqrt{0.0383}\sqrt{10}\cdot 10^{6} \frac{m}{s}\sim 619278\frac{m}{s}\sim 619.3\frac{Km}{s} [/tex] Using more precise values for radius ad Sun's mass will give about [itex] 617.54\frac{Km}{s} [/itex]. Daniel. PS.I don't know how u got that figure under the radical. |
| Jan2-05, 01:06 PM | #3 |
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Thnx. I see what youve done but don't understand why. How come you broke it down to 2 factors:
[tex]v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}[/tex] And how come you multiplied by 4 instead of 2? Thnx |
| Jan2-05, 01:27 PM | #4 |
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Recognitions:
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Escape velocity
Here's the easy way to calculate it using google calculator
The mistake in your calculation is that the value of G you use is in MKS units G = 6.67300 × 10^-11 m^3 kg^-1 s^-2 and you put in the radius of the sun in km, not meters. |
| Jan2-05, 03:27 PM | #5 |
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[tex] \sqrt{ab}=\sqrt{a}\sqrt{b} [/tex] ,i could break it into 2 square roots,putting the powers of 10 aside. That '4' appears from the product between 2 (initially in the formula) and 2 (from the Sun's mass). Daniel. |
| Jan2-05, 03:40 PM | #6 |
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Ahh, ok 1.9891
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