Physics - Classical Mechanics - Gravitational Potential Energy

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Introduction

Hey it's a me again @drifter1! In this article we will continue with Physics, and more specifically the branch of "Classical Mechanics". Today's article will be about Gravitational Potential Energy. So, without further ado, let's get straight into it!


Gravitational Potential Energy

In the Work and Energy chapter we saw the usefulness of Work and Potential.

Work W is defined as:

The measure of how much energy is being transferred by one object when this object is moved over a distance by an external force which is applied in the direction of displacement.

Potential Energy (in general) can be thought of as:

The energy that is stored on a system.

Potential Energy U only depends on the relative position of the various parts of a system and the initial and final configuration of those parts.

We also saw that Gravitational Potential Energy is defined as:



where:

  • U : Gravitational Potential Energy (in Joules)
  • w : Force of Gravity (Weight) (in Newtons)
  • h = h2 - h1 : Height Difference (in meters)
  • m : Mass of the attracted object (in Kg)
  • g : Gravitational Acceleration (in m/s2)
But, how do we end up with this equation?

Proof

The Gravitational Potential Energy is equal to the work done by the Force of Gravity, while moving a mass m from height h1 to h2:



The force Fh is considered to be pointing from the Earth's center towards the surface (inverse) making the value negative:



We want to end up with W = U1 - U2, which means that we define U as:



Using this definition ΔU = U2 - U1 will be negative when h1 > h2, which is what we expected.

In gravity problems we will put U = 0 at the best possible point, so that calculations become easier.

By combining the equation of U with the equation of Gravitational Acceleration g we end up with:


Gravitational Potential

The Gravitational Potential V is defined as the Gravitational Potential Energy U that a unit mass m possesses:



The value of V is always negative and maximum at infinity. The S.I. unit of V is the J/Kg.


From the Earth to the Moon (Escape Velocity)

Using the Law of Conservation of Energy, we can easily calculate the velocity at which we have to be launched at to escape the Earth's Gravity and reach the moon.

Let's suppose that the final velocity is 0 and that U2 = 0, meaning that the height is thought as infinite.



This velocity is called escape velocity.

In the case of the Earth the value is 11.186 Km/s.


RESOURCES:

References

  1. https://opentextbc.ca/universityphysicsv1openstax/chapter/13-3-gravitational-potential-energy-and-total-energy/
  2. https://phys.libretexts.org/Courses/University_of_California_Davis/UCD%3A_Physics_9A__Classical_Mechanics/7%3A_Gravitation/7.3%3A_Energy_in_Gravitational_Systems
  3. https://byjus.com/jee/gravitational-potential-energy/
  4. http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

Images

Mathematical equations used in this article, where made using quicklatex.


Previous articles of the series

Rectlinear motion

Plane motion

Newton's laws and Applications

Work and Energy

Momentum and Impulse

Angular Motion

Equilibrium and Elasticity

Gravity


Final words | Next up

And this is actually it for today's post!

Next time we will start getting into Exercises around Newtonian Gravity (I'm currently thinking of 2 parts)

See ya!

Keep on drifting!
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