## Friday, September 20, 2019

### Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Energy...

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

Gravitational Energy Conservation

While moving an object from a distance r1 to a distance r2 from the center of gravity, the gravitational field has performed certain work W[r1,r2], spending certain amount of energy. Since energy must be conserved, it should materialize in some other way.

Indeed, the kinetic energy of a probe object at the end of its movement from point {r1,0,0} to {r2,0,0} must be equal to the work performed by the field.

We have positioned our probe object at point {r1,0,0} without any initial speed, that is Vr1=0. Therefore, the kinetic energy Kr1 at this initial point is zero.
At the end of a motion at point {r2,0,0} the speed Vr2 must have such a value that the kinetic energy Kr2 would be equal to work W[r1,r2] performed by the field.
Kr2 = m·V²r2 /2 = W[r1,r2]

From this equation, knowing how to calculate the work performed by a gravitational field (see the previous lecture), we can find a speed of a probe object at the end of its motion from point {r1,0,0}, where it was at rest, to point {r2,0,0}:
m·V²r2 /2 =
(G·M /r2 − G·M /r1
)·m
r2 = 2·G·M·(1/r2 − 1/r1)

In a particular case, when r1=∞ and r2=r, that is a probe object falls with no initial speed from the infinitely long distance from a source of gravity to a point at distance r from it, the formula is simplified:
r = 2·G·M /r

We would like to warn against falling into a point-mass that is a source of gravity, when the final distance from it is zero, that is r=0 in the above equation. It obviously produces infinite speed and infinite kinetic energy, which does not correspond to reality. The most important reason for this deviation from the reality is our assumption about a source of gravity to be a point-mass. Real objects have certain non-zero dimensions. For example, in case of a gravitational field around our planet should not be analyzed by this formula for values of r less than the radius of Earth.

Back to energy conservation.
The potential energy of an object is a measure of work that it can do, if left alone, that depends on a position of an object relative to other objects and such properties as its mass. Actually, these two parameters are the only ones needed to calculate the potential energy of a probe object in a gravitational field, provided we know everything about the field.

As we know (see the previous lecture), amount of work we need to move a probe object of mass m from an infinite distance to a distance r from a source of gravity equals to
Wr = −G·M·m /r
It's negative from our external to the gravitational field viewpoint, because we don't actually perform work, the field performs it for us. So, from the external viewpoint, the field gives certain energy to external object by performing some work on it, similar to a person, pushing the cart, spends energy, transferring it to a cart.

In this expression, skipping over the universal gravitational constant G, components M (mass of a source of gravitational field) and r (distance from the center of the gravitational field) characterize the gravitational field, while m (mass of a probe object) characterizes the object, whose potential energy we measure.

This energy is transferred to a probe object as its potential energy. If an object is not moving from this position, because some force holds it there, it retains this potential energy. As soon as there is no force holding it there, it will start moving towards the source of gravity, losing its potential energy and gaining the kinetic energy because it will move faster and faster.

As an example, let's calculate the kinetic energy and final speed of a small asteroid, free falling on the surface of the Moon, assuming the Moon is the only source of gravity in the Universe.
The Universal Gravitational Constant is
G=6.67408·10−11,
its units are m3·kg−1·sec−2.
The mass of the Moon is M=7.34767309·1022 kg.
The radius of the Moon is r=1.7371·106 m.
Let's assume that an asteroid falling on the Moon is relatively small one, say, m=50 kg.

According to the formula above, the gravitational field of the Moon did the work that equals to
W ≅ 6.67408·10−11·7.34767309·1022·50 /(1.7371·106) ≅ 141,151,800 (joules)
let's check the units to make sure we get joules, the units of work
m3·kg−1·sec−2·kg·kg·m−1 = kg·m2·sec−2 = N·m = J

The final speed V can be calculated by equating this amount of work and kinetic energy of an asteroid:
V2 ≅ 2·6.67408·10−11·7.34767309·1022 /(1.7371·106) ≅ 5646072
let's check the units to make sure we get the square of speed units
m3·kg−1·sec−2·kg·m−1 = m2·sec−2 = (m/sec)2

From this the speed of an asteroid falling from infinity onto Moon's surface is
V ≅ √5646072 ≅ 2376 (m/sec)

Incidentally, this is the so-called escape speed from the Moon, the initial speed needed for an object to leave the gravitational field of the Moon. A stone, thrown perpendicularly to the surface of the Moon with an initial speed less than that will go for certain distance away from the Moon, but then it will be brought back by the Moon's gravitation. Only if the initial speed is equal or exceeds the one above, the distance an object will go will be infinite, that is the object will leave the gravitational field of the Moon.

Let's do similar calculations for the Earth, using the same assumptions, the same asteroid and the same units of measurement.
M = 5.972·1024 kg
r = 6.371·106 m
W ≅ 6.67408·10−11·5.972·1024·50 /(6.371·106) ≅ 3,128,049,424 (joules)

V2 ≅ 2·6.67408·10−11·5.972·1024 /(6.371·106) ≅ 125121977 (m/sec)2
From this the speed of an asteroid falling from infinity onto Earth's surface is
V≅√125121977≅11186 m/sec