A particle of mass m is projected with a velocity
By BYJU'S Exam Prep
Updated on: September 13th, 2023
Given, that a particle of mass m is projected with a velocity v = kvₑ from the surface of the earth.
vₑ = escape velocity
We have to find the maximum height above the surface reached by the particle.
The principle of conservation of mechanical energy states that “The total mechanical energy of a system is conserved” i.e., the energy can neither be created nor be destroyed; it can only be internally converted from one form to another if the forces doing work on the system are conservative in nature.
In hydroelectric power plants, waterfalls on the turbines from a height which in turn, rotates the turbines and generates electricity.
Through the conservation of mechanical energy,
1/2 mv2 – GmM/R = -GmM/(R+h)
v2/2 – GM/R = -GM/(R+h)
v2/2 = GM/R – GM/(R+h)
v2/2 = GM/R(1 – 1/h)
v2/2 = GMh/R(R+h)
Now, 1/2 k2vₑ2 = GMh/R(R+h)
We know, that vₑ = √2GM/R
So, vₑ2 = 2GM/R
Now, 1/2 k2(2GM/R) = GMh/R(R+h)
Canceling out common terms, we get
k2 = h/(R+h)
k2(R+h) = h
Rk2 + hk2 = h
Rk2 = h – hk2
Rk2 = h(1-k2)
h = Rk2/(1-k2)
Therefore, the maximum height above the surface reached by the particle is Rk2/(1-k2).
Summary:
A particle of mass m is projected with a velocity v =kve(k<1) from the surface of the earth. (ve = escape velocity). The maximum height above the surface reached by the particle is – R is the radius of the earth. (a). Rk2/1-k2 (b). R(k/1-k)2 (c). R(k/1+k)2 (d). R2k/1+k
A particle of mass m is projected with a velocity v =kve(k<1) from the surface of the earth. (ve = escape velocity). The maximum height above the surface reached by the particle is Rk2/(1-k2).
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