Solving Escape Velocity of the Sun: Troubleshooting an Error

DB
Messages
501
Reaction score
0
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.

v_e=\sqrt{\frac{2GM}{r}} or v_e=\sqrt{\frac{2\mu}{r}}

v_e_\odot=617.54(km/s)

v_e_\odot=\sqrt{\frac{2*6.673e-11*1.9891e30(kg)}{696000(km)}}

v_e_\odot=\sqrt{~3.8141e14}\neq 617.54(km/s)

What am I doing wrong?
Thanks
 
Physics news on Phys.org
Okay,i'll make some approximations,but i'll get the result and the order of magnitude.
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}

Using more precise values for radius ad Sun's mass will give about 617.54\frac{Km}{s}.

Daniel.

PS.I don't know how u got that figure under the radical.
 
Thnx. I see what youve done but don't understand why. How come you broke it down to 2 factors:

v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}

And how come you multiplied by 4 instead of 2?
Thnx
 
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.
 
DB said:
Thnx. I see what youve done but don't understand why. How come you broke it down to 2 factors:

v_{esc.,Sun}\sim\sqrt{\frac{4\cdot6.673}{696}}\sqrt{\frac{10^{30}10^{-11}}{10^{6}}}\frac{m}{s}

And how come you multiplied by 4 instead of 2?
Thnx

Since
\sqrt{ab}=\sqrt{a}\sqrt{b}
,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.
 
Ahh, ok 1.9891
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top