**SfC Home > Physics > Gravitation >**

# Effect of Sun on Escape Velocity from Earth

by Ron Kurtus (revised 8 March 2011)

The escape velocity equation allows you to calculate the velocity an object—such as a rocket—must attain in order to completely overcome the gravitational pull of the Earth. However, the equation does not take into consideration the effect of the Sun's gravitation on the escape velocity.

When a rocket blasts off from the Earth in a direction away from the Sun, it must not only escape the Earth's gravitation but also the gravitational pull from the Sun. Surprisingly, the escape velocity from the Sun at the Earth's surface is greater than the escape velocity from the Earth. The two factors must be combined to give the true or total escape velocity.

Questions you may have include:

- What is the standard escape velocity from Earth?
- What is the effect of the Sun on the escape velocity?
- What is the total escape velocity?

This lesson will answer those questions. Useful tool: Units Conversion

## Escape velocity from Earth

The escape velocity from Earth can be calculated from the equation:

v_{E}= − √(2GM/R_{i})

where

**v**is the escape velocity from Earth in km/s_{E}**G**is the Universal Gravitational Constant = 6.674*10^{−20 }km^{3}/kg-s^{2}**M**is the mass of the Earth = 5.974*10^{24}kg**R**is the initial separation between the centers of the Earth and the rocket_{i}

Note: The negative sign in the equation indicates that the velocity vectoris in a direction opposite of the gravitational force vector, according to our direction convention.

If the mean radius of the Earth is 6371 km and the rocket shut its engines at 334.4 km (such as with the Saturn V rocket), the initial separation from the Earth's center would be:

R6371 km + 334.4 km = 6705.4 km_{i}=(

See Gravitational Escape Velocity with Saturn V Rocket for more information.)

Substituting values into the equation results in:

v2*(6.674*10_{E}= − √[^{−20})*(5.974*10^{24})/(6705.4)]km/s

v118.920_{E}= − √()km/s

v10.905 km/s_{E}= −

A rocket would have to achieve this velocity before shutting off its engines, if it were to escape from the gravitational pull of the Earth.

## Escape velocity from Sun at Earth

Suppose a rocket blasted off the Earth from its far side, away from the Sun.

Rocket leaves Earth away from Sun

The escape velocity from the Sun would be:

v_{S}= − √(2GM_{S}/D)

where

**M**is the approximate mass of the Sun = 1.988*10_{S}^{30}kg**D**is the approximate separation between the Sun and the rocket on the far side of the Earth = 1.496*10^{8}km

Note:The separation between the center of the Sun and the center of the Earth is approximately 1.496*10^{8}km. The addition of the 6705.4 km would make a negligible contribution.

Substituting values into the equation results in:

v2*6.674*10_{S}= − √(^{−20}*1.988*10^{30}/1.496*10^{8})km/s

v17.738*10_{S}= − √(^{2})km/s

v42.1 km/s_{S}= −

The escape velocity from the Sun at the Earth's surface is larger than the escape velocity from the Earth itself. This means that the rocket may be able to escape the Earth but would have to go much faster to escape the Sun's gravitation.

## Combined escape velocity

If the rocket exceeded the escape velocity from the Earth but not the escape velocity from the Sun, it would move off into space and then soon reverse directions and fall into the Sun.

The combined escape velocity from the Earth, adding in the effect from the Sun, is:

vkm/s_{e}= − √(v_{E}^{2}+ v_{S}^{2})

v10.9_{e}= − √(_{}^{2}+ 42.1_{}^{2})km/s

v1891.22_{e}= − √(^{})km/s

v43.488 km/s_{e}= −

This velocity does not take into account the rotation of the Earth and its orbital velocity, which will affect the escape velocity. This is beyond the scope of our studies. Also, the contribution from the gravitation of the Moon is negligible.

### Moving toward the Sun does not apply

The escape velocity concept fails when the rocket blasts from the Earth on the side facing the Sun, because rocket is no longer escaping the Earth's gravitation and moving out to infinity. Instead, the rocket is leaving the Earth and being attracted to the Sun.

## Summary

The escape velocity from the Earth does not take into account the escape velocity from the Sun that an object—such as a rocket—must attain in order to completely overcome the gravitational pull of both the Earth and the Sun.

When a rocket blasts off from the Earth in a direction away from the Sun, you must combine the escape velocity from Earth and the escape velocity from the Sun to get the true escape velocity.

Shoot for the stars in your efforts

## Resources and references

### Websites

**Acceleration due to Gravity Calculations** - from Western Washington University

### Books

**Top-rated books on Gravitation**

## Questions and comments

Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.

## Share

Click on a button to bookmark or share this page through Twitter, Facebook, email, or other services:

## Students and researchers

The Web address of this page is:

**www.school-for-champions.com/science/
gravitation_escape_velocity_sun_earth.htm**

Please include it as a link on your website or as a reference in your report, document, or thesis.

## Where are you now?

## Effect of Sun on Escape Velocity from Earth