# Homework Help: Escape speed in space

1. Apr 28, 2005

### Honore

QUESTION:
A space station orbits the sun at the same distance as the earth but on the opposite side of the sun. A small probe is fired away from the station. What minimum speed (m/s) does the probe need to escape the solar system?

MY UNDERSTANDING AND SOLUTION:
The escape speed v from a sphere of radius R and mass M is given by the energy-conservation equation as follows: (from "Schaum's 3000 Solved Problems in Physics" book, page 101)

(1/2)*m*v^2 = G*M*m / R where;

M: mass of the sun (=1.98*10^30 kg)
m: mass of the small probe
R: Radius of the sun (=6.95*10^8 m)
G: Universal Gravitation Constant [=6.67*10^(-11) Nm^2/kg^2]

From the equation typed in bold above;

v = sqrt(2*G*M / R) and using the numerical values v is found about

616479 m/s .

What do you think?

Thanks.

2. Apr 28, 2005

### DaveC426913

Six hundred kilometers per second???

Does that sound reasonable???

Without examining too closely, one thing I note is that you haven't calc'd the escape velocity from Earth orbit, you've calc'ed the escape velocity from the surface of the Sun: R = 695,000km. R should be Earth's orbit.

3. Apr 28, 2005

### Honore

Okay, let us take R as the distance between the sun and earth since the problem says "A space station orbits the sun at the same distance as the earth but on the opposite side of the sun", then v is found

Is this reasonable? Our professor says yes, but it is incorrect. So what?

4. Apr 28, 2005

### BobG

Are you sure your numbers were right? You're in the ballpark, but definitely wrong.

The heliocentric gravitational constant (GM) is 1.32712442076 x 10^11 km

One astronomical unit (average radius of the Earth's orbit) is 1.49597871 x 10^8 km

Just eyeballing it tells you your escape velocity will be a little less than the square root of 2000, which puts you in 40's of km/sec.

5. Apr 13, 2010

### Student003

vescape = [2*G*M/R].5

be sure to use:
G = 6.67x10-11
M = (mass of the sun) = 1.99x1030 kg
R = (mean disntance from the earth to the sun) = 1.5x1011 meters

that should give you the right answer

6. Apr 13, 2010

### Student003

by the way, the answer should be 4.21x104 m/s

7. Apr 13, 2010

### DaveC426913

That's great but if he's still struggling with this after five years, he's got bigger problems...