You are a venture capitalist that is asked to invest in a startup company that claims it will be able to launch tiny "micro space probes" into space at close to the speed of light using a massive electromagnetic rail gun system2. You are cynical about their cost estimates and decide to analyze the problem in more detail before you invest in their company. Neglect air resistance for this worksheet. 1. A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?
2. Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?
43. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.
4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the railgun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).
a. Will the recoil momentum of the ship be relativistic? Justify your argument.
b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?

1. 5.825 × 10¹⁷ J

2. i. $ 29.125 billion ii. I would not invest in the company

3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. a i. The momentum will not be relativistic

ii. This is because objects with large masses do not move at relativistic speeds

b i. 155 m/s

ii. This speed wouldn't be a problem for the ship.

Explanation:

1.  A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?

The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)

So, K = (γ - 1)mc²

= (1/√(1 - β²) - 1)mc²

= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²

= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²

= 11.65 × 10¹⁶ kgm²/s²

= 1.165 × 10¹⁷ J

Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have

20 % of E = K

0.2E = K

E = K/0.2

= 1.165 × 10¹⁷ J/0.2

= 5.825 × 10¹⁷ J

2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?

i. how much would this launch cost?

Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion

ii. Would you invest in this company?

I would not invest in the company

3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.

Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy,  then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).

a. Will the recoil momentum of the ship be relativistic? Justify your argument.

i. Will the recoil momentum of the ship be relativistic?

The momentum will not be relativistic.

ii. Justify your argument.

This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic

b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?

i. At what speed will the ship recoil after it launches a probe?

Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J

Also, E'² = (pc)² + (mc²)² where p = momentum of payload

So, making p subject of the formula, we have

(pc)² = E'² - (mc²)²

pc = √[E'² - (mc²)²]

p = √[E'² - (mc²)²]/c

substituting the values of the variables into the equation, we have

p = √[E'² - (mc²)²]/c

p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s

p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s

p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s

p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s

p = 18.59 × 10¹⁶/3 × 10⁸ m/s

p = 6.20 × 10⁸ kgm/s

From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.

So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship

So, v = p/m'

= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg

= 1.55 × 10² m/s

= 155 m/s

ii. Do you think that this is a problem for the ship?

Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.

the independent variable, or the one that is intentionally manipulated is the speed.

it's a b and