# Gravitational Potential Energy

## Practice

### practice problem 1

- Calculate the speed needed to escape the Earth from its surface.
- Calculate the speed needed to escape the Sun from the Earth's orbit.
- Calculate the speed needed to escape the Milky Way from our solar system. (The sun is 25,000 light years from galactic center — halfway to the edge of the galaxy. Its orbit encloses a mass that is 78 billion times the mass of the Sun.)

#### solution

A quick review of the origin of the escape velocity equation. Start with the law of conservation of energy.

*E*_{0} = *E*

The relevant energies are kinetic and gravitational potential.

*K*_{0} + *U*_{0} = *K* + *U*

Replace the energy symbols with their equations.

½m_{1}v_{0}^{2} − |
Gm_{1}m_{2} |
= ½m_{1}v^{2} − |
Gm_{1}m_{2} |

r_{0} |
r |

Let *r* → ∞ and *v* → 0.

½m_{1}v_{0}^{2} − |
Gm_{1}m_{2} |
+ = 0^{} − 0 |

r_{0} |

Get rid of the subscripts and the minus sign.

½v^{2} = |
Gm |

r |

Solve for *v*.

v = √ |
2Gm |

r |

Substitute appropriate numbers and calculate the speed needed to…

escape the Earth.

*v*= √2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(5.97 × 10 ^{24}kg)6.37 × 10 ^{6}m*v*= 11,200 m/sFast!

escape the Sun.

*v*= √2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(1.99 × 10 ^{30}kg)1.50 × 10 ^{11}m*v*= 42,100 m/sDamn fast!

escape the galaxy.

*v*= √2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(78 × 10 ^{9})(1.99 × 10^{30}kg)(25,000)(365.25×24×60×60) (3.00 × 10 ^{8}m/s)*v*= 296,000 m/sUnbelievably fast!

### practice problem 2

- If the Sun were a black hole what would be the radius of its event horizon?
- If the Earth were a black hole what would be the radius of its event horizon?
- If you were a black hole what would be the radius of your event horizon?

#### solution

A quick review of the origin of the event horizon equation. Start with the law of conservation of energy.

*E*_{0} = *E*

The relevant energies are kinetic and gravitational potential.

*K*_{0} + *U*_{0} = *K* + *U*

Replace the energy symbols with their equations.

½m_{1}v_{0}^{2} − |
Gm_{1}m_{2} |
= ½m_{1}v^{2} − |
Gm_{1}m_{2} |

r_{0} |
r |

Let *v*_{0} = *c*, *r* → ∞ and *v* → 0.

½m_{1}c^{2} − |
Gm_{1}m_{2} |
+ = 0^{} − 0 |

r_{0} |

Get rid of the subscripts and the minus sign.

½c^{2} = |
Gm |

r |

Solve for *r*.

r = |
2Gm |

c^{2} |

Substitute appropriate numbers and calculate the radius of…

the event horizon of the Sun.

*r*=2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(1.99 × 10 ^{30}kg)(3.00 × 10 ^{8}m/s^{2})^{2}*r*= 2950 mI could easily walk this distance in under an hour — even if I doubled it to get the diameter. That's not far. Could the dying sun actually collapse to this size? Highly unlikely.

the event horizon of the Earth.

*r*=2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(5.97 × 10 ^{24}kg)(3.00 × 10 ^{8}m/s^{2})^{2}*r*= 8.884 mmI love how close that is to 9 mm. It makes me see a 9 mm bullet in my mind. But this 9 mm event horizon is a radius and the 9 mm for bullets is a diameter. Doesn't really matter much since this is a hypothetical calculation. Crushing the Earth down to the point where it would fit into the barrel of any conventional weapon is totally absurd.

the event horizon of you (or a person somewhat like you).

*r*=2(6.67 × 10 ^{−11}N m^{2}/kg^{2})(70 kg) (3.00 × 10 ^{8}m/s^{2})^{2}*r*= 1 × 10^{−25}mWhat is the standard person? Most of the serious readers of this book (well, this website claiming to be a book) are probably young adults. Older adults don't have much reason to study physics. (How sad.) Let's give them a characteristic mass of 70 kg, because why not. (That's not your mass, but it's certainly within an order of magnitude of your's.)

Why am I having this internal monolog with you? Your mass is 70 kg because I said so and your event horizon is ten orders of magnitude smaller than the radius of a proton (

*r*~ 10^{−15}m). Squeeze a human down to the size of a proton and they still won't become a black hole. You need to go ten billion times smaller.

### practice problem 3

When we look at galaxies and other objects outside our own Milky Way we see that they are generally moving away from us and that their recessional velocities are nearly directly proportional to their distance. This observation was first made in 1929 by the American astronomer Edwin Hubble (1889–1953) and is now known as Hubble's law. Mathematically, Hubble's law is written as…

*v* = *Hr*

where…

v = |
the object's recessional velocity (usually stated in km/s) |

r = |
the object's distance from the Milky Way (usually stated in megaparsecs or Mpc) |

H = |
a constant of proportionality known as the Hubble constant (69.3 ± 0.8 km/s/Mpc). |

Perform the following set of calculations.

- Given that one parsec equals 3.08568 × 10
^{16}m…- Convert the Hubble constant to SI base units.
- By how much does a meter of space expand in ten years? About how big is this?
- How far away in light years is a distant quasar if it appears to be moving away from us at 90% of the speed of light?
- How far away in light years is the edge of the observable universe? (Your answer to this question can also be used to determine the age of the universe.)

- Will the universe continue expanding forever or will gravity eventually cause it all to collapse in a big crunch?
- Derive an expression for the critical density of the universe. (Hint: Use the formulas for escape velocity, the Hubble law, and the density of a uniform sphere.)
- Compute the critical density in terms of hydrogen atoms per cubic meter.
- Speculate on the fate of the universe given that the mean density of a galaxy is roughly one hydrogen atom per cubic centimeter while the mean density of the space between galaxies is about one hydrogen atom per cubic meter.
- Speculate on the fate of the universe given that the Hubble constant appears to be increasing.

#### solution

The solutions to this practice problem can be found in the discussion section of this topic.

### practice problem 4

Signatories to the Outer Space Treaty of 1967 agreed that they "shall not place nuclear weapons or other weapons of mass destruction in orbit or on celestial bodies or station them in outer space in any other manner". That hasn't stopped them from doing feasibility studies, however, or looking for loopholes in the law. Is a massive object a weapon? What if there was a satellite orbiting the Earth that was full of massive objects? Massive objects with dimensions similar to a telephone pole made of a dense material? Would that be a weapon? Drop one from space and tell me what happens.

Schemes like this have been in the works in the United States since 1964 (under the informal name of Project Thor) and are revived from time to time — for example, by the RAND Corporation in 2002 and by the US Air Force in 2003. They have been described as hypervelocity rod bundles, orbital telephone poles, and most poetically rods from God. They have met with legal, political, economic, and scientific skepticism. They were also used as a plot device in the 2013 action-adventure movie GI Joe: Retaliation.

characteristic | value |
---|---|

platform altitude | 8000 km |

rod diameter | 40 cm |

rod length | 7 m |

rod material | tungsten carbide |

rod density | 15,630 kg/m^{3} |

Given the values in the table, determine the following quantities for a hypothetical, hypervelocity, orbiting, rod bundle system…

- the orbital
*speed*of the platform - the orbital
*period*of the platform - the mass of one rod
- the energy to de-orbit one rod (the energy needed to change its orbital speed to zero)
- the gravitational potential energy of one rod relative to the surface of the Earth
- in joules
- in tons of TNT equivalent

- the impact velocity of a de-orbitted rod (disregarding aerodynamic drag and the rotational motion of the Earth)

#### solution

The orbital speed of the platform

*F*=_{c}*F*_{g}*mv*^{2}= *Gm*_{1}*m*_{2}*r*^{2}*r**v*= √*Gm**r**v*= √(6.67 × 10 ^{−11}N m^{2}/kg^{2})(5.97 × 10 ^{24}kg)(6,670,000 m + 8,000,000 m) *v*= 5210 m/sThe orbital period of the platform in hours and minutes

∆ *t*=∆ *s*= 2π *r**v**v*∆ *t*=2π(6,670,000 m + 8,000,000 m) 5,210 m/s ∆ *t*= 17,692 s = 4:55The mass of one rod

*m*= ρ*V*= ρπ*rh**m*= (15,630 kg/m^{3})π(0.20 m)(7 m)*m*= 68,700 kg = 68.7 tonnesThe energy to de-orbit one rod (the energy needed to change its orbital speed to zero)

*K*= ½*mv*^{2}*K*= ½(68,700 kg)(5,210 m/s)^{2}*K*= 9.32 × 10^{11}J = 932 GJThe gravitational potential energy of one rod relative to the surface of the Earth

*U*= −_{g}*Gm*_{1}*m*_{2}*r*∆ *U*= −_{g}*Gm*_{1}*m*_{2}⎛

⎜

⎝1 − 1 ⎞

⎟

⎠*r*_{0}*r*∆ *U*=_{g}(6.67 × 10 ^{−11}N m^{2}/kg^{2})(5.97 × 10 ^{24}kg)(68,700 kg)× ⎛

⎜

⎝1 − 1 ⎞

⎟

⎠(6,670,000 m) (6,670,000 m + 8,000,000 m) ∆

*U*= 2.24 × 10_{g}^{12}J = 2240 GJ = 2.24 TJOne ton of TNT is equivalent to 4.184 GJ by definition.

∆ *U*=_{g}2240 GJ 4.184 GJ/ton TNT ∆

*U*= 535 tons TNT_{g}I would call this half a ton of TNT to be less precise, but more meaningful.

The impact velocity of a de-orbitted rod (disregarding aerodynamic drag and the rotational motion of the Earth)

*v*= √2∆ *U*_{g}*m**v*= √2(2.24 × 10 ^{12}J)68,700 kg *v*= 8075 m/sDamn, that's fast.