Gravitational Potential Energy

Okay, in class today, Mrs. K handed back our Transparency Worksheet. After that, we looked over the notes she handed out yesterday.

Gravitational Potential Enegy
From the past unit, we learned two formulas for Potential Energy:
Ep=mgh and Es=1/2kx²

Here's another one that we learned from class today.
PEg= -(Gm₁m₂)/R

From the Newton's Law of Universal Gravitation Fg= Gm₁m₂/R, we increases the separation distance from R₁ to R₂. It requires work and when it is done, the gravitational PE increases.

Derivation of PEg formula:
PEg = ((-Gm₁m₂)/R₂) - ((-Gm₁m₂)/R₁)

Gravitational Potential Well
- two objects that has force of attraction between them having negative potential energy.
- it should rise out of the Earth's potential well to be free of the gravitational force.

For example: If one of the masses is the Earth then the other mass is on Earth's ground level, the total energy is just the PEg since KE = 0 J.

Total Energy = KE + PEg

If the object rises from the Earth's ground level, it would have both PEg and KE, so then, the total energy = KE + PEg.

Escape Velocity
- the minimum velocity of an object to escape from Eath's potential well.

To calculate escape velocity, KE = - PEg but we take the absolute value of the PEg to have a positive velocity so then it will be KE = |-PEg|.

let's say m₁=mass of the object, v₁=velocity of the object, m2=Earth's mass, R=Earth's radius, G=gravitational constant.
KE= 1/2 m₁v₁²
PEg= - Gm₁m₂/R

1/2 m₁v₁² = Gm₁m₂/R

v₁ = √(2Gm₂)/R (m₁ will reduce)

Total Energy and Binding Energy
Binding Energy - amount of additional kinetic energy an object needs to escape.
- it is equal to the negative of gravitational potential energy

on Earth's ground level Et = - Gm₁m₂/R
so, Eb = Gm₁m₂/R

if an object is in orbit at any radius R₁, Fc keeps the object in circular orbit and Fg provided the force attraction between the object and Earth.
Fc=Fg
m₁v₁²/R₁ = Gm₁m₂/R₁² (R₁² will reduce to R₁)
m₁v₁² = Gm₁m₂/R₁

the Et of the orbiting object is
Et = KE + PEg
Et = 1/2 m₁v₁² + (-Gm₁m₂/R)

since m₁v₁² = Gm₁m₂/R₁

Et = 1/2 Gm₁m₂/R₁ + (-Gm₁m₂/R)
Et = -1/2 Gm₁m₂/R₁ or Et = 1/2 PEg

so then the binding energy is Eb = 1/2 Gm₁m₂/R₁

After the discussions, she handed out a worksheet called 'Gravitational Potential Energy Questions'.
Here are the solutions.

1.) PEg = - Gm₁m₂/R
= (- (6.67 * 10^-11)(500)(5.98 * 10²⁴))/ (6.37 * 10⁶)
= -3.13 * 10⁶ J

2.) Et = Ek + Ep
= 1/2 m₁v₁² + ( - Gm₁m₂/R₁)
= - 1/2 (Gm₁m₂/R₁)
= - 1/2 ((6.67 * 10^-11)(5.98 * 10²⁴)(500))/ (4.22 * 10⁷)
= - 2.36 * 10⁹ J

3.) Etave. = Et(in orbit) - Et (on Earth)
= -2.36 * 10⁹ - (-3.13 * 10¹⁰)
= 2.89 * 10¹⁰ J

4.) Eb = 2.36 * 10⁹ J is needed to remove satellite from Earth's orbit.

5.) Ep = - Gm₁m₂/R
= (- (6.67 * 10^-11)(5.98 * 10²⁴)(2000))/ (6.38 * 10⁶ + 400 * 10³)
= -1.18 * 10¹¹ J

Before the end of class, she handed out notes about Escape Speed and asked us to do Questions 28,29,30,31,32,33 in the duck book. I guess this is the end of my post, I hope i covered everything.

Next scribe is Russel L.