# Escape velocity

• Jun 28th 2008, 11:35 AM
pfcbias
Escape velocity
I am concerned on what is the velocity from the Sun at the distance of 1000 AU
• Jun 28th 2008, 01:00 PM
topsquark
Quote:

Originally Posted by pfcbias
I am concerned on what is the velocity from the Sun at the distance of 1000 AU

$v_{esc}= \sqrt{\frac{GM_S}{r}}$

We have that r = 1000 AU and we need to convert that to m. So
$\frac{1000~AU}{1} \cdot \frac{149597870691~m}{1~AU} = 149597870691000~m \approx 1.496 \times 10^{14}~m$

And we know that
$M_S = 1.989 \times 10^{30}~kg$

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

-Dan
• Jun 28th 2008, 02:32 PM
pfcbias
Astronomy
(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 PM
wingless
Quote:

Originally Posted by pfcbias
(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?

Hints:
1. Angular velocity doesn't change.
2. $F =m\cdot a$
3. $G = m\cdot g$
• Jun 29th 2008, 01:24 AM
CaptainBlack
Quote:

Originally Posted by pfcbias
(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?

Quote:

Originally Posted by wingless
Hints:
1. Angular velocity doesn't change.
2. $F =m\cdot a$
3. $G = m\cdot g$

The answer in red is wrong.

Kepler's third law applies:

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

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

Translating this into terms of angular velocities gives:

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

where $\omega$ denotes an angular velocity

RonL
• Jun 29th 2008, 01:30 AM
wingless
Quote:

Originally Posted by CaptainBlack
The answer in red is wrong.

Kepler's third law applies:

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

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

Translating this into terms of angular velocities gives:

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

where $\omega$ denotes an angular velocity

RonL

I thought your answer too. But the question doesn't state orbiting, which confused me. I may have misunderstood that.
• Jun 29th 2008, 02:14 AM
CaptainBlack
Quote:

Originally Posted by wingless
I thought your answer too. But the question doesn't state orbiting, which confused me. I may have misunderstood that.

Circluar speed at 1AU only has meaning for an orbit.

RonL