I am concerned on what is the velocity from the Sun at the distance of 1000 AU

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- Jun 28th 2008, 11:35 AMpfcbiasEscape velocity
I am concerned on what is the velocity from the Sun at the distance of 1000 AU

- Jun 28th 2008, 01:00 PMtopsquark
$\displaystyle v_{esc}= \sqrt{\frac{GM_S}{r}}$

We have that r = 1000 AU and we need to convert that to m. So

$\displaystyle \frac{1000~AU}{1} \cdot \frac{149597870691~m}{1~AU} = 149597870691000~m \approx 1.496 \times 10^{14}~m$

And we know that

$\displaystyle M_S = 1.989 \times 10^{30}~kg$

Now plug in the numbers. For reference I get $\displaystyle v_{esc} = 942~m/s$

-Dan - Jun 28th 2008, 02:32 PMpfcbiasAstronomy
(Wondering)Thanks alot,

Just 3 more questions....

1. If circular speed at 1 AU from the Sun is 30 km/s. What is the circular speed at 20 AU?

2. The acceleration of a 4 kg body to which an unbalanced 18 N force is applied

3. The acceleration of gravity on Mars is 0.38 Earth gravities. How much would a dog that weighs 80 N on Earth weigh on Mars?

- Jun 28th 2008, 03:33 PMwingless
- Jun 29th 2008, 01:24 AMCaptainBlack
The answer in red is wrong.

Kepler's third law applies:

$\displaystyle \frac{\tau_1^2}{\tau_2^2}=\frac{r_1^3}{r_2^3}$

where $\displaystyle \tau$ is the period and $\displaystyle r$ the radius of an orbit.

Translating this into terms of angular velocities gives:

$\displaystyle \frac{\omega_2^2}{\omega_1^2}=\frac{r_1^3}{r_2^3}$

where $\displaystyle \omega$ denotes an angular velocity

RonL - Jun 29th 2008, 01:30 AMwingless
- Jun 29th 2008, 02:14 AMCaptainBlack