Escape Velocity Formula
Everything you need to know about the escape velocity formula is provided below. Please proceed to read the whole document carefully to understand the topic completely.
When we toss anything up in the air, it does not travel through the air and into space. The gravitational force is the reason behind this. So, how can a rocket reach the furthest reaches of the universe? To break beyond the earth’s gravitational pull, the spacecraft requires a massive amount of fuel. This clarifies what the earth’s escape velocity or escape speed is.
The object must attain the escape velocity of celestial bodies such as planets and natural satellites. This permits the celestial body to be free of the gravitational sphere’s influence. In this case, the system’s total kinetic energy and gravitational potential energy will be zero.
The least velocity at which a body must be thrown to transcend the earth’s gravitational attraction is called escape velocity. It is the speed at which an object must travel to escape the gravitational field; i.e., to leave the land without ever falling back. An object with this velocity at the earth’s surface will completely escape the gravitational field of the planet, even if losses due to the atmosphere are considered.
A spaceship leaving Earth’s surface, for example, must travel at 7 miles per second, or roughly 25,000 miles per hour, to avoid colliding with the surface.
Formula For Escape Velocity
Escape velocity formula is given by
V = 2GMR
V is the escape velocity
G is the gravitational constant is 6.67408 × 10-11 m-3kg-1 s-2
M is the mass of the planet
R is the radius from the centre of gravity
- An alternative expression for escape velocity particularly useful at the surface of the body is
Vesc = 2gR
- Where g is the acceleration due to the gravity of the earth.
- It is expressed in m/s and the escape velocity of earth is 11,200 m/s.
- The escape velocity formula is applied in finding the escape velocity of any body or any planet if mass and radius are known.
Determine the escape velocity of Jupiter if its radius is 7149 Km and mass is 1.898 × Kg
Given: Mass M = 1.898 × Kg,
Radius R = 7149 Km
Gravitational Constant G = 6.67408 ×10-11 m-3kg-1 s-2
Escape Velocity is given as
Ves = 2GMR
=√2 x 6.67408 × 10-11 x 1.898 × / 7149
Determine the escape velocity of the moon if Mass is 7.35 × 1022 Kg and the radius is 1.5 × m.
M = 7.35 × 1022 Kg,
R = 1.5 × 106 m
Escape Velocity formula is given by
Vesc = 2GMR
= √2×6.673××7.35×1022 / 1.5×
= 7.59 × m/s