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Escape Velocity Formula

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TL;DR
The escape velocity formula does not depend on the escaping object’s properties. The only thing that counts is the celestial body’s mass and radius. It is given by the relation
ve=2GMR


You’ve probably seen rockets launch into space. These rockets need a huge kick or push to leave the earth’s surface. This push is needed because of the earth’s strong gravitational field. So here comes to escape velocity.

Escape Velocity Formula

In this article learn about the escape velocity formula and how to use this formula to find the escape velocity of any object or body. Before going any further let us discuss What is escape velocity.

What is Escape Velocity

You might have noticed a ball thrown up in the air. When we throw it upwards it ascends to a certain height and falls back to the Earth. If we again throw it with greater velocity than the previous then, in this case, it rises to a greater height than the previous launch.
Can we keep on increasing the velocity of the ball to move to a height from where it does not return back to earth? Is it possible? And the answer is yes it is possible.

So,

If an object is launched from the Earth’s surface at a high enough velocity so that it can ascend into space and never return. This is known as the Earth’s escape velocity.

Now that we have defined escape velocity let us look at its formula.

For any massive celestial body, the escape velocity formula or equation is given by the relation

ve=2GMR

where,

ve is escape velocity

M is the mass of the planet or celestial body

R is the radius of the planet and

G is a Gravitational constant and its value is given by G=6.67×1011Nm2Kg2

from this escape velocity formula, we can clearly see that escape velocity does not depend on the properties of the escaping object.
The mass and radius of the planet or celestial body in question are the only factors that matter.

The escape velocity of the earth is 11.2 km/s which corresponds to about 25,000 miles per hour.

Now that we’ve learned about the escape velocity formula, we’ll look at some problems that use it.

Some Important Escape Velocities

The table given below lists the escape velocities of some celestial bodies that you might want to remember as they are often used while solving problems related to this formula.

Celestial BodyEscape Velocity
Moon2.3 Km/s
Mercury4.28 Km/s
Earth11.2 Km/s
Jupiter60 Km/s
Sun618 Km/s
Neutron Star2×105 Km/s

Escape Velocity Formula – Solved Example Problems

Question 1. Calculate a body’s escape velocity from the moon. Assume that the moon is a uniform sphere with a radius of 1.76×106m and a mass of 7.36×1022Kg.

Value of universal gravitational constant is G=6.67×1011Nm2Kg2

Solution.

Let us first write down the information given in the question. It is given that

Radius R=1.76×106m,

Mass m=7.36×1022Kg

Escape velocity is given by the relation

ve=2GMR

putting in all the values we gat

ve=2×6.67×1011×7.36×10221.76×106

calculating it we get

ve=2375m/s=2.375Km/s

Question 2. A black hole is a body from which nothing can escape. What conditions must be met for a uniform spherical body of mass M to be a black hole? What should the radius of a black hole be if it has a mass nine times that of the Earth? (Mass of Earth ME=6×1024Kg and G=6.67×1011Nm2Kg2 )

Solution.

Even light should not escape a body that is a black hole. This is because from Einstein’s special theory of relativity we know that speed of any object can not exceed the speed of light (c=3×108m/s). So, c is the upper limit to the projectile’s escape velocity. Hence for the body to be a black hole,

ve=2GMRc

Now if the mass of the black MB hole is nine times then the mass of earth ME

MB=9×ME

putting in the value of the mass of earth we get

MB=9×6×1024Kg

Now by rearranging escape velocity formula for the value of R we get

R=2GMc2=2×6.67×1011×9×6×1024(3×108)2

on calculating we get

R8×102m or nearly 8 cm.

Question 3. Jupiter has 318 times the mass of the Earth, and its radius is 11.2 times that of the Earth. Calculate the escape velocity of a body from Jupiter’s surface assuming that the escape velocity from Earth’s surface is 11.2 Km/s.

Solution.

Escape velocity from the surface of Earth is

ve=2GMERE=11.2Km/s

where, ME is mass of Earth and RE is radius of Earth

ve=2GMJRJ

where, MJ is mass of Jupiter and RJ is radius of Jupiter and ve is the escape velocity from Jupiter.

Now it is given in the question that

MJ=318ME and RJ=11.2RE

therefore,

ve=2G(318×MR)11.2×RE=2GMERE×(31811.2)=ve×31811.2=11.2×31811.2=59.7Km/s

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