Answer the following question with full explanation : The escape velocity of a body from the surface of the earth is 45.5 km/s. If a satellite were to orbit close to the earth's surface, what would be it's critical velocity.
Solved questions according to escape velocity .Solve the question according to the escape velocity.
Answers
Explanation:
The escape velocity of a body from the surface of the earth is 45.5 km/s. If a satellite were to orbit close to the earth's surface, what would be it's critical velocity.
Solved questions according to escape velocity .Solve the question according to the escape velocity.
Solution : Here, a/c to given data;
➜ v(escape) = 45.5 km/s
➜ v(escape) = 45.5 * 10³ m/s
Escape velocity : The minimum velocity with which object (body) should be projected from the earth's surface, so that it can be escape from the earth's gravitational field.
Now, by formula;
➜ v(escape) = √2 * v(critical)
Hence, the critical orbital speed of a satellite in a near earth orbit is given by
➜ v(critical) = v(escape)/√2
➜ v(critical) = (45.5 * 10³)/√2
➜ v(critical) = {45.5/√2} * 10³
➜ v(critical) = {(45.5 * √2)/2} * 10³
➜ v(critical) = {(45.5 * 1.414)/2} * 10³
➜ v(critical) = {64.337/2} * 10³
➜ v(critical) = 32.1685 * 10³ m/s
Answer : Hence, critical velocity of the satellite is 32.1685 * 10³ m/s.
[Note : 1/√2 = (1/√2) * (√2/√2) = √2/(√2 * √2) = √2/2]