Effect of Sun on Escape Velocity from Earth
About 3 min read
(updated 30 May 2023)
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:
vE = − √(2GM/Ri)
where
- vE is the escape velocity from Earth in km/s
- G is the Universal Gravitational Constant = 6.674*10−20 km3/kg-s2
- M is the mass of the Earth = 5.974*1024 kg
- Ri is the initial separation between the centers of the Earth and the rocket
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:
Ri = 6371 km + 334.4 km = 6705.4 km
Substituting values into the equation results in:
vE = − √[2*(6.674*10−20)*(5.974*1024)/(6705.4)] km/s
vE = − √(118.920) km/s
vE = − 10.905 km/s
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.
The escape velocity from the Sun would be:
vS = − √(2GMS/D)
where
- MS is the approximate mass of the Sun = 1.988*1030 kg
- D is the approximate separation between the Sun and the rocket on the far side of the Earth = 1.496*108 km
Substituting values into the equation results in:
vS = − √(2*6.674*10−20*1.988*1030/1.496*108) km/s
vS = − √(17.738*102) km/s
vS = − 42.1 km/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:
ve = − √(vE2 + vS2) km/s
ve = − √(10.92 + 42.12) km/s
ve = − √(1891.22) km/s
ve = − 43.488 km/s
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