**Evans SPH 4U1**

**Physics Grade 12**

**Unit 4:
Energy & Momentum**

**Note 7:
More on Gravitational Potential Energy**

Reference: Chapter 6.3

*Recall:*

Any two masses have a
Gravitational Potential Energy of
at a
separation distance of r.

As discussed earlier, it is negative because of the reference we use
where earth is in a potential well.

*Refer to Figure 4 (Text
page 288)*

If a rocket is at rest on the surface of the earth its kinetic energy is zero. If the rocket is given a speed at the surface of the earth, it now has kinetic energy and begins to rise. As it rises its potential energy increases (becomes less negative) and it will reach a point where its potential energy is at its maximum and its kinetic energy is zero. At this point gravity will pull the rocket back to earth . When it hits the earth it has the same kinetic energy it left with (i.e. total energy is constant).

There is a speed the rocket
can have to escape the potential well of Earth (i.e. escape
gravitational force). This is when the rocket's kinetic energy equals
the depth of the potential well at the Earth's surface (when the total
energy is 0). *Derive this speed. 11.2 km/s*

If a rocket's total energy is negative then it will not be able to escape from earth's potential well.

**Escape Speed:** The minimum
speed needed to project a mass to just escape the gravitational force.

**Escape Energy:** The minimum
kinetic energy needed to project a mass to just escape the
gravitational force.

**Binding Energy: **The amount
of additional kinetic energy needed by a mass m to just escape from a
mass M.

general formula for binding energy

**Kinetic Energy of a
satellite in circular orbit **-derive:

**Total energy of a
satellite in circular orbit** (it** **is negative and is equal
to): - derive

**Binding energy for a
satellite bound to earth **is equal to:

*It is important for you
to understand the difference between binding energies and total
energies of a stationary object in space and an object which is in
orbit.*

**Event Horizon: **The
surface of a black hole

**Example 1:**

A 60.0 kg space probe is in
a circular orbit around Europa. If the orbital radius is 2.00 x 10^{6}m
and the mass of Europa is 4.87 x 10^{22}kg, determine the:

a) kinetic energy of the
probe and it orbital speed (Answer: 4.87 x 10^{7} J; 1.27 x 10^{3}
m/s)

b) gravitational potential
energy of the probe (Answer: -9.74 x 10^{7} J)

c) total orbital energy of
the probe (Answer: -4.87 x10^{7} J)

d) binding energy of the
probe (Answer: 4.87 x 10^{7} J)

e) additional speed that the probe must gain in order to break free of Europa (Answer: 528 m/s)

f) the energy needed to
move the probe (from the 2.00 x 10^{6}m orbital radius) to an orbital
radius of 4.00 x 10^{6}m above Europa (Answer: 2.43
x 10^{7}J)

**Suggested Text
Questions:**

See sample problem #2 page 291

Page 294 # 6-8

Page 300 Review #'s 10-12, 17