## Additional Velocity Required to Overcome Earth's Gravitational Pull for Satellite Escape

Escape velocity is the minimum velocity an object must attain to break free from the gravitational attraction of a massive body without further propulsion. For an object on the surface of the Earth, this velocity is approximately 11.2 kilometers per second (or about 25,000 miles per hour). But for a satellite already in orbit, the calculation is little more complex.

Satellites in low Earth orbit (LEO) typically travel at speeds around 8 kilometers per second (km/s). This velocity, known as orbital velocity, is necessary to counteract the gravitational force pulling the satellite towards Earth, allowing it to maintain a stable orbit.

However, what if we seek to free a satellite from Earth’s gravitational grasp altogether? Escaping Earth’s gravitational influence requires imparting additional velocity to the satellite. Suppose we have a satellite orbiting near the Earth’s surface, with an orbital velocity of approximately 8 km/s. To break free from Earth’s gravity, an additional velocity of approximately 3.2 km/s is needed.

## Example

### Q.A spaceship is launched into a circular orbit close to the earth's surface. What additional velocity has now to be imparted to the spaceship in the orbit to overcome the gravitational pull. Radius of earth =6400 km , g= 9.8 ms^{-2}

### (A) 3.2 km/s

### (B) 11.2 km/s

### (C) 1.5 km/s

### (D) 8 km/s

#### Table of Contents

## Solution

###### We know that , orbital speed of satellite is

###### v_o=\;\sqrt{\frac{GM}r}\;

###### where\;r\;is\;dis\tan ce\;from\;centre\;of\;planet/earth

###### Also, r= R+ h where R is radius of planet/earth and h is height of satellite from surface of earth

###### v_o=\;\sqrt{\frac{GM}{R+h}}\;

###### Near to earth's surface, h can be neglected

###### So, R\;+\;h\;\approx\;R

###### v_o=\;\sqrt{\frac{GM}{R}}\;

###### v_o=\;\sqrt{\frac{gR^2}{R}}\;

###### v_o=\;\sqrt{gR}\;

###### v_o\;=\;\sqrt{Rg}\approx\;8\;km/s\;

###### v_e\;=\;\sqrt{2Rg}\approx\;11.2\;km/s\;

###### Addition velocity required is

###### v_{add}\;=\;v_e\;\;-\;v_{o\;}

###### v_{add}\;=11.2 - 8 = 3.2 km/sec

###### (A) is correct option

## Conclusion

This additional velocity is crucial for overcoming the gravitational potential energy barrier that binds the satellite to Earth. When the satellite reaches this escape velocity, its kinetic energy surpasses the gravitational potential energy, allowing it to break away from orbit and venture into interplanetary space.

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