# Is v = 11,2km/s enough to wrest away from the Sun's gravitation

1. Apr 8, 2016

### wiegetz

1. The problem statement, all variables and given/known data
Are 11,2km/s enough, to wrest away from the Suns gravitational field? (We are departing away from the Earth planet, away from the sun)

Here I am unsure if I should use Suns Radius or Earths Radius r. I think that Suns Radius would be more logical, since the problem is about getting away from the Gravitational Field of the Sun. Might be wrong though :)

The Formula I need for this: v = √2γM/r
Suns Mass: 1,989*1030kg
Suns Radius: 696.342 km = 696342000m
(Earths Radius : 6370km = 6370000m
2. Relevant equations
v = √2γM/r

3. The attempt at a solution
v = √2γM/r

Inserting the Data : v = √2*6,672*10-11*m3*s-2*1,989*1030kg/696342000m

My Result: 617375,1049m/s or ≈ 617,4km/s is needed to wrest away from Suns gravitational field.

Solution: No, 11,2km/s is far not enough to wrest away from Suns gravitational field.

2. Apr 8, 2016

### Staff: Mentor

You don't want the Sun's radius, you want the radial distance from the center of the Sun of your starting location.

Things are made a bit more complicated if you are departing from the surface of the Earth, since you effectively have to escape from both the Sun and the Earth. You can assume that the effects of the other solar system bodies are negligible. If you're starting from the Earth's surface then you cannot ignore its effect since its close proximity makes its influence significant.

3. Apr 8, 2016

### Staff: Mentor

617 km/s is necessary if you start from the surface of sun.
Unless you start from the surface of something, the radius of objects is completely irrelevant.

4. Apr 8, 2016

### wiegetz

Thanks for the answers!
The teacher said that this is quite simple and that we only need to google up the mass of Sun.
About the radial distance : I am sure that you are correct, but the teacher would mention it if we needed that as well. Is there no other possiblity to calculate it, starting from earth?

Radial Distance Earth-Sun = 149597870000m

My solution if I use the radial distance:

v = √2*6,672*10-11*m3*s-2*1,989*1030kg/149597870000m

42120,90592m/s = 42,1 km/s

Last edited: Apr 8, 2016
5. Apr 8, 2016

### Staff: Mentor

Same as what?

Your calculated value of 42.1 km/s is good for an escape from a point located at Earth's distance from the Sun. That is, a location that is the same distance from the Sun as the Earth but without the Earth being there too. But it's not enough if you are departing from the Earth's surface. Then you wold need to escape both bodies' pull.

6. Apr 8, 2016

### wiegetz

Answer same as followed: Solution: No, 11,2km/s is far not enough to wrest away from Suns gravitational field.

We did not talk about bodies' pull at all. Do you think that it would be OK to use 42.1 km/s as a result then?

7. Apr 8, 2016

### PeroK

... you've also got the initial orbital speed of the Earth around the Sun to take into account!

8. Apr 8, 2016

### wiegetz

Might be correct, but that question is supposed to be really simple, as the teacher told.

I repeat again: The only thing he told is that we need to google up the Suns mass, nothing more.

9. Apr 8, 2016

### Staff: Mentor

Then you've shown that this is the correct answer. You need greater than 42 km/s in order to escape the Sun from the vicinity of the Earth's orbit.

10. Apr 8, 2016

### wiegetz

Solved, thanks for the help!