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Escape Velocity from Gravity
by Ron Kurtus (updated 29 May 2023)
The escape velocity from gravity is the initial velocity of an object that is projected upward such that it is sufficient to overcome the downward pull of gravity and not allow the object to fall back to the ground.
Note: Some textbooks call it escape speed, but that term is misleading and incorrect, since speed can be in any direction, while velocity is a vector that requires a direction. Escape velocity is in a direction away from the Earth.
Also, since the object is at or near the surface of the Earth, it is often called surface escape velocity.
The equation for the escape velocity comes from the gravitational escape velocity equation, as applied to objects near the Earth's surface. The resulting simple equation gives the escape velocity as a function of the acceleration due to gravity and the Earth's radius. You can also apply the equation to the Moon and Sun, provided you know the radius and gravity of each of those bodies.
Although the values for the surface escape velocity are commonly used in textbooks and scientific papers, they are unfortunately misleading and not realistic. There are several problems that prevent the escape velocity from gravity from being practical.
Questions you may have include:
- How do you determine the escape velocity from gravity?
- What are the escape velocities for the Earth, Moon and Sun?
- What are problems with the equation near Earth?
This lesson will answer those questions. Useful tool: Units Conversion
Determining escape velocity from gravity
The escape velocity from gravity can be obtained from the general gravitational escape velocity equation by substituting in the acceleration due to gravity.
This requires that the object is projected upward at some initial velocity from relatively close to the Earth's surface—typically, less than 65 km or 40 miles in altitude. The equation does not apply to continuously propelled objects, such as rockets. Also, air resistance and the rotation of the Earth are not considered.
It is sometimes called surface escape velocity, since the escape velocity occurs at or very near the surface of the Earth.
Gravitational escape velocity equation
The gravitational escape velocity equation is:
ve = −√(2GM/R)
where
- ve is the escape velocity in kilometers/second (km/s)
- G is the Universal Gravitational Constant (6.674*10−20 km3/kg-s2)
- M is the mass of the planet or sun in kilograms (kg)
- R is the separation between the center of mass of the planet or sun and the center of the object in kilometers (km)
Note: The negative sign in the equation indicates that the velocity vector is in a direction opposite of the gravitational force vector, according to our direction convention. Be aware that many textbooks state the equation as a positive value.
(See Convention for Direction in Gravitation Equations for more information.)
Also note: Since escape velocity is typically stated in km/s, the value of the Universal Gravitation Equation, G, is defined in km3/kg-s2 and the separation R in km.
(See Gravitational Escape Velocity Derivation for more information.)
Relationship with acceleration due to gravity
For objects on the surface of the Earth, the acceleration due to gravity is:
g = GME/RE2
where
- g is the acceleration due to gravity = 9.8 m/s2 or 0.0098 km/s2
- ME is the mass of the Earth = 5.974*1024 kg
- RE is the radius of the Earth = 6371 km
Note: Since escape velocity is in km/s, g is stated in km/s2 instead of m/s2.
Also note: The value of g is considered constant at heights near the Earth's surface. At an altitude of about 64 km or 40 mi, the value of g reduces 2%, going from 9.8 m/s2 to 9.6 m/s2 and the assumption of being constant starts to fail.
(See Gravity Constant for more information.)
Equation for escape velocity from gravity
To find the equation for the escape velocity from gravity, first multiply both sides of the relationship by RE:
gRE = GME/RE
Substitute into ve = − √(2GM/R) to get the equation for the escape velocity from gravity:
ve = − √(2gRE)
Note: The velocity is a negative number to indicate it is moving in the opposite direction of gravity.
Factors for escape velocity from Earth's gravity
Note that ve would have to be about 11.2 km/s or 25,055 miles per hour. It is unrealistic for an object to reach that speed at or even near the surface of the Earth.
In other words, the calculated escape velocity from 191 km above the Earth's surface is about 39,685 km/hr or 24,684 mi/hr. 11.023 km/s
It would take the rocket until 191 km to get to the required speed.
Escape speed
Although ve is a vector that indicates the direction of the velocity as away from the Earth, it is more convenient to simply indicate the magnitude or speed of the escape velocity. Thus, the equation you usually see is:
se = √(2gRE)
where se is the escape speed or magnitude of the escape velocity vector ve.
The escape from gravity equation is simpler to use than the gravitation equation, especially since the mass of the Earth is not necessary in calculations.
Escape velocity at an angle
If you project an object straight up, it will travel until it reaches its maximum height and then fall back to the ground. If you project it at an angle, it will follow a parabolic path until it hits the ground.
If the horizontal velocity is sufficient, the object will go into orbit around the Earth. If the vertical velocity is sufficient, the object will follow a curved path but never return to Earth.
(The lesson on Gravity and Newton's Cannon goes through the various possibilities.)
In other words, the angle does not affect the value of the escape velocity from gravity.
Common escape velocities
You can use the equation to determine escape velocity from the gravity of the Earth, Moon and Sun.
Earth
The radius of the Earth, RE, is about 6371 km and g = 0.0098 km/s2. The escape velocity or speed away from the Earth's surface is:
se = √(2gRE)
se = √[2*(0.0098 km/s2)*(6371 km)]
se = √(124.872 km2/s2)
se = 11.175 km/s
Thus, the escape velocity from the surface of the Earth is about 11.2 km/s or 25,055 miles per hour.
This is the same value calculated from se = − √(2GME/RE) at the Earth's surface.
Moon
The radius of the Moon is RM = 1737 km and the acceleration due to the Moon's gravity is gM = 0.00162 km/s2. Its surface escape velocity is:
se = √(2gMRM)
se = √[2*(0.00162 km/s²)*(1737 km)]
se = √(5.63) km/s
se = 2.4 km/s
This is the same as the gravitational escape velocity at the surface. However, the effects of the gravitational pull of the Earth and Sun on the escape velocity have not been taken into consideration.
Sun
The radius of the Sun is about RS = 6.955*105 km and its acceleration due to gravity is gS = 0.274 km/s2. The escape velocity from the Sun's gravity is:
se = √(2gSRS)
se = √[2*(0.274 km/s²)*(6.955*105 km)]
se = √(38.113*104) km/s
se = 617.4 km/s
Typically, atomic particles escaping the Sun in a solar storm would reach such great altitudes to escape that the vs= √(2gSRS) equation would not be valid.
Problems with equation near Earth
The calculated escape velocity from gravity near the Earth's surface of 11.2 km/s or 26,000 miles per hour is too high to be practical. Also, the effect of the Sun is not taken into account.
Assumes extremely high acceleration
A major problem with the escape velocity from gravity value is that the velocity is calculated at or near the Earth's surface. An infinite acceleration would be required to project an object at 11.2 km/s from the Earth's surface.
Also, it would be very difficult—if not impossible—for a rocket to attain a velocity of 11.2 km/s relatively close to the Earth's surface. The Saturn rocket that was used to go to the Moon did not reach that speed until it was over 193 km (180 miles) from the Earth's surface.
(See Gravitational Escape Velocity with Saturn V Rocket for more information.)
Escape velocity value becomes inaccurate
At the 193 km altitude, the gravity escape velocity equation is inaccurate. Instead of being 11.2 km/s, the value for se using the gravitation escape velocity equation is 11.007 km/s.
You could get that value from the gravity escape velocity equation, provided you recalculated g for that altitude and added 193 km to the radius. However, it would be better to simply use the gravitation escape velocity equation.
Rocket would burn up
Also, in order to reach the escape velocity at lower altitudes, the rocket would be traveling at hypersonic speed, which would be so far above the speed of sound that it could cause the burn-up of a rocket exterior before it left the Earth's atmosphere.
Realistically, a rocket would have to build up its speed until it reached the extreme upper atmosphere, where air resistance is negligible at high speeds.
Factor of the Sun
Adding to the escape velocity from the Earth is the gravitational pull from the Sun. This is another factor that must be considered, requiring the gravitational escape velocity equations and not the gravity equations.
(See Effect of Sun on Escape Velocity from Earth for more information.)
Equation not practical
Although the se = √(2gRE) equation gives a correct value of the escape velocity near Earth, it really is more of an approximation for actual applications.
Note: It pains me to see this value being so carelessly used. It is just unscientific.
Summary
The velocity an object projected upward from the Earth must attain such that it will overcome the gravitational pull and not fall back to the ground can be approximated by the equation se = √(2gRE), provided the object starts near the Earth's surface. A similar equation can be used for the Moon, Sun and other celestial bodies.
However, the equation is not applicable for a rocket blasting off from Earth, due to the extreme acceleration required, effects of the atmosphere and the gravitation from the Sun. Instead, the gravitational escape velocity equation should be used.
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Escape Velocity from Gravity