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.

Test Scribe

Hey guys! Sorry for the late post. In today's class we just spent the whole period writing the Dynamic Test. And that's it. Have a good night everyone!

Next scribe is RoselS.