## Tuesday, April 2, 2019

### Unizor - Physics4Teens - Energy - Potential Energy - Gravity

Notes to a video lecture on http://www.unizor.com

Energy of Gravity

An object of mass m is raised on certain height h above the surface of the Earth.
Analyze its potential energy, as a function of height h.
Do not assume that the force of gravity is constant, but rather use the Newton's Law of Universal Gravitation.
Assume that the radius of Earth is R and its mass is M.

Solution:

If x is the distance from the center of the Earth to an object, the force of gravity equals to
F(x) = G·M·m/
(where G is a gravitational constant approximately equaled to 6.674·10-11N·m2·kg−2).

Potential energy of an object at height h above the ground is the work performed by the force of gravity as an object falls down to the surface of the Earth from the distance R+h from the center to the distance R.

Since the force of gravity varies, we should use the calculus to calculate this work.
An infinitesimal increment of work dW(x), assuming the force F(x) acts at a distance dx, equals to
dW(x) = F(x)·dx

Integrating this by x from R+hto R, we get the full work and the potential energy of an object at height h above the ground:
Epot = W(h) = RR+hF(x)·dx =
= G·M·m·
[1/R − 1/(R+h)]

Obviously, when an object is at the ground level (h=0), its potential energy equals to zero and, as it moves higher (h is increasing), its potential energy grows to its maximum value of G·M·m/R, as the object moves farther and farther from the ground.

When the height of the object his significantly smaller than the radius of the Earth (h << R), our formula for potential energy can be approximated with a simpler expression:
Epot = W(h) =
= G·M·m·
[1/R − 1/(R+h)] =
= G·M·m·h/
[R·(R+h)] ≅
≅ G·M·m·h/R² = m·g·h
,
where g=G·M/ is the free fall acceleration at the ground level, approximately equaled to
g ≅ 9.81 m/sec²
and P = m·g is the weight of an object at the ground level, assumed constant for small heights h.
In this case the potential energy is simply a product of weight (force of gravity) and distance - a classical expression for work performed by a constant force:
Epot = W(h) ≅ P·h

Let's also analyze the kinetic energy of the object.
Firstly, we assume that our object starts its free falling from the height H above the ground with zero initial velocity.

Its potential energy at any height h we know from the above calculation:
Epot = G·M·m·[1/R − 1/(R+h)]
Its kinetic energy depends on its speed V:
Ekin = m·V²/2
The problem is, we don't know how the speed depends on the height h.

What we do know is the dependency of the force of gravity F on the height h:
F = G·M·m/(R+h)²
We also know that speed V is the first derivative of height hby time t and the acceleration is the second derivative:
V(t) = h'(t)
a(t) = h"(t)
Therefore, according to the Newton's Second Law,
F = m·a = G·M·m/(R+h)²
Now we have an expression for acceleration a:
a = G·M/(R+h)²
Let's underscore that acceleration is the second derivative of distance h, as a function of time t:
h"(t) = G·M/[R+h(t)]²

We will not attempt to solve this differential equation to get a function h(t), take its derivative to get the speed and substitute into a formula for kinetic energy.
Instead, we will apply a clever trick.
Let's multiply the last expression by 2h'(t):
2h'(t)h"(t) =
= 2G·M·h'(t)/
[R+h(t)]²
The reason for this operation is the following.
The left side of this equation is the derivative of the square of the speed:
d/dt{[h'(t)]²} = 2h'(t)·h"(t)
and on the right side of this equation we find another derivative:d/dt{2G·M/[R+h(t)]} =
= 2G·M·h'(t)/
[R+h(t)]²
Therefore, we have the equality of the derivatives:
d/dt{[h'(t)]²} =
d/dt{2G·M/[R+h(t)]}

If derivatives are equal, the functions differ by a constant:
[h'(t) = 2G·M/[R+h(t)+ C
The constant C can be determined, using the conditions at the beginning of motion at time t=0:
h(0) = H and h'(0) = 0
Therefore,
[h'(0) = 2G·M/[R+h(0)+ C
[0 = 2G·M/[R+H+ C
C = − 2G·M/[R+H]
Now we can write
[h'(t) =
= 2G·M/
[R+h(t)]−2G·M/[R+H]
From this we can express the kinetic energy
Ekin = m·V²(t)/2 =
= m·
[h'(t)]²(t)/2 =
= G·M·
{1/[R+h(t)]−1/[R+H]}

Since our task was to express the energy in terms of height above the ground, we can simply write it as
Ekin = G·M·[1/(R+h)−1/(R+H)]

Finally, let's find the full mechanical energy of an object falling from the sky.
Epot = G·M·m·[1/R−1/(R+h)]
Ekin = G·M·[1/(R+h)−1/(R+H)]
Efull = Epot + Ekin =
= G·M·
[1/R−1/(R+H)]
As we see, the full mechanical energy is constant, it does not depend on the height. As an object falls from the sky, its potential energy is decreasing, but kinetic energy increasing, and the sum of both types of energy remains constant.