# Escape velocity

1. Jan 2, 2005

### DB

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

2. Jan 2, 2005

### dextercioby

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.

3. Jan 2, 2005

### DB

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

4. Jan 2, 2005

### pervect

Staff Emeritus
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.

5. Jan 2, 2005

### dextercioby

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.

6. Jan 2, 2005

### DB

Ahh, ok 1.9891