How is escape velocity independent of angle of projection??
I mean i expect it to be DEPENDENT on the angle of projection cuz only vertical component of initial velocity helps a body to overcome gravitational force. So only the vertical component of velocity can behave as escape velocity. And angle does matter there.. PLzzzz help me out..
Assuming there is no atmosphere and the projectile is not aimed below the horizon (i.e. at the ground), the escape velocity IS independent of launch angle. Consider the situation from an energy point of view. If there is enough in initial kinetic energy so that Ve^2/2 = MG/R, where M is the Earth's mass and R is its radius, then when r = infinity, the total energy
(m/2)v^2 - mMG/r
is zero, and the projectile can escape. The horizontal component of velocity at launch IS used to overcome gravitational attraction at large distances.
break components of velocity and apply conservation of energy . You will observe that sin( theta ) and cos(theta) terms of velocity can be easily eliminated from sin square + cos square = 1. Hence we see that expression of escape velocity of object launched at angle theta from vertical will be same as that for vertical projection of body.
The term "escape velocity" refers to the minimum velocity an object needs to escape the gravitational pull of a celestial body, such as a planet or a moon. It is indeed independent of the angle of projection. Let me explain why.
When an object is launched into space, it follows a curved path due to the gravitational force acting on it. This path is known as a projectile trajectory. The trajectory is dependent on the initial velocity and launch angle, and it can be decomposed into horizontal and vertical components.
The vertical component of the initial velocity is responsible for overcoming the gravitational force pulling the object downwards. The object's velocity decreases as it moves upward due to the gravitational force acting against it, eventually reaching zero at the maximum height. From that point on, the gravitational force begins to accelerate the object back towards the celestial body, causing its velocity to increase in the downward direction.
However, if the object had enough initial velocity in the vertical direction to reach zero velocity at the maximum height, it would continue moving upward and eventually escape the gravitational pull. This is the point where the escape velocity comes into play.
Now, let's examine the role of the launch angle. The launch angle determines the initial direction of the projectile trajectory. If the object is launched at an angle other than vertical (e.g., an upward or downward angle), its initial velocity will still have vertical and horizontal components. The vertical component influences the object's ability to overcome gravity, while the horizontal component determines the range and direction of its path.
When calculating escape velocity, we consider a purely vertical launch, as we are interested in the minimum velocity to escape gravity. By doing this, we ignore the horizontal component and focus solely on the vertical component.
So, regardless of the angle of projection, the escape velocity remains the same. It is solely determined by the mass of the celestial body and its radius, according to the equation:
ve = √(2GM/r)
Where:
- ve is the escape velocity
- G is the universal gravitational constant
- M is the mass of the celestial body
- r is the distance from the object's center to the body's center (radius)
Therefore, the escape velocity is independent of the angle of projection, as it depends solely on the mass and radius of the celestial body.