Rocket Escape Velocity from the Earth-Sun system

Click For Summary
The discussion focuses on calculating the energy conservation equation for a rocket escaping the Earth-Sun system. The key point is determining the rocket's position after it is freed from Earth's gravity, which is considered to occur at "infinity" from Earth. However, this infinity is not negligible when measured from the Sun due to the Sun's dominant mass. The gravitational potential energy can be treated additively, allowing calculations to be performed in two steps, but it's noted that the escape process occurs simultaneously from both bodies. Additionally, the Earth's motion relative to the Sun must be factored into the calculations.
Rikudo
Messages
120
Reaction score
26
Homework Statement
Earth with mass m orbits the sun (mass M) in circular path (radius R) with velocity v. If a rocket is launched from the earth,find the initial velocity that is needed so that it could leave the solar system (free from both the sun's and earth's gravity).

Note-
this process undergoes two independent steps: free from the earth's gravity, and then free from the sun's gravity.
Relevant Equations
-
I have a difficulty when making the energy-conservation-equation for the second step.

When making the equation, we need to know the exact position (measured from the sun) of the rocket after it is freed from the Earth gravitation.

But, where exactly does the rocket free from Earth gravitation? All I know is that this happens at infinity (measured from the earth).

Does this "infinity" negligible if we measure it from the sun? Why?
 
Physics news on Phys.org
The Earth's mass is much smaller than the Sun, so you could approximate a long way from Earth as still a distance ##R## from the Sun.
 
  • Like
Likes Rikudo and malawi_glenn
Alternately, one could realize that gravitational potential energy is additive. If you add the gravitational potential energy of the rocket in the field of the Earth to the gravitational potential energy of the rocket in the field of the Sun the result is the gravitational potential energy of the rocket in the field of the Sun and Earth combined.

So you can do the calculations in two separate steps and add the results together. But there is no need to do the physical thrusting in two distinct steps. You are actually escaping from both bodies at the same time.
 
In addition to what has already been mentioned regarding adding two speed delta's being a reasonable approximation, you should also remember to consider that the Earth already moves relative to the Sun.
 
If have close pipe system with water inside pressurized at P1= 200 000Pa absolute, density 1000kg/m3, wider pipe diameter=2cm, contraction pipe diameter=1.49cm, that is contraction area ratio A1/A2=1.8 a) If water is stationary(pump OFF) and if I drill a hole anywhere at pipe, water will leak out, because pressure(200kPa) inside is higher than atmospheric pressure (101 325Pa). b)If I turn on pump and water start flowing with with v1=10m/s in A1 wider section, from Bernoulli equation I...

Similar threads

  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K