Question

The electronic systems on the New Horizons spacecraft, launched on January 19,2006 and scheduled to reach Pluto on July 14 , 2015, are powered by electricity generated by heat. The heat comes from the radioactive decay of \({ }^{233}\) Pu in the \(11 \mathrm{~kg}\) of \(^{238} \mathrm{PuO}_{2}\) fuel onboard. The generator provided \(240 \mathrm{~W}\) when the spacecraft was launched. If the power output is directly proportional to the amount of \({ }^{233}\) Pu in the generator, what will the power output be when the spacecraft reaches Pluto? The half-life of \({ }^{238} \mathrm{Pu}\) is \(87.7 \mathrm{y}\).

Short answer

About 235.878 watts.

Step-by-step solution

Understand the Proportionality

The power output of the generator is directly proportional to the amount of \ \(^{238}\text{Pu}\) remaining. This means that the power decreases as the amount of \ \(^{238}\text{Pu}\) decays over time. We will need to calculate the remaining amount of \ \(^{238}\text{Pu}\) using its half-life.

Calculate Time Elapsed

Calculate the total time from the launch to when the spacecraft reaches Pluto: 2015 - 2006 = 9 years.

Apply the Half-Life Formula

The formula to calculate the remaining amount after a certain time is given by \[ N(t) = N_0 \times (\frac{1}{2})^{\frac{t}{T_{1/2}}} \]where - \ N_0 \ is the initial quantity of the substance,- \ t \ is the time elapsed,- \ T_{1/2} \ is the half-life of the substance.Substitute the given values: \[ t = 9 \text{ years, } T_{1/2} = 87.7 \text{ years} \]

Calculate Remaining Fraction of \(^{238} \mathrm{Pu}\)

Using Step 3's formula, \[ N(t) = N_0 \times (\frac{1}{2})^{\frac{9}{87.7}} \]Calculate the fraction of \ \(^{238}\text{Pu}\) left.

Compute the Remaining Power Output

Assume initial power is \( 240 \, W \). The fraction of \ ^{238} Pu \ remaining after 9 years is applied to this:\[ P(t) = 240 \, W \times (\frac{1}{2})^{\frac{9}{87.7}}\]Calculate \ P(t) \, which gives the power output when the spacecraft reaches Pluto.

Key concepts

Plutonium-238

Plutonium-238 ( ^{238} ext{Pu} ) is a radioactive isotope of plutonium commonly used as a heat source in space missions. This isotope is particularly valued for its ability to produce heat through radioactive decay, which can be converted into electricity via thermoelectric generators. This makes it a vital component in spacecraft that need a reliable power source far from the Sun. Unlike solar power, Plutonium-238 can provide continuous power in the dark expanses of space, enabling long-duration missions.

One of the benefits of Plutonium-238 is its relatively long half-life, which ensures that it provides power for extended periods without requiring much maintenance or replacement. This property makes it particularly suitable for missions like New Horizons, as it can still power its systems even after years of travel in space.

Half-Life Calculation

The concept of half-life is fundamental in understanding radioactive decay. It refers to the time it takes for half of a radioactive substance to decay into another element or isotope. For Plutonium-238, the half-life is 87.7 years.

To determine how much of a substance is left after a certain period, we use the half-life formula:\[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]- N_0 is the initial amount of the substance.
- t is the time elapsed.
- T_{1/2} is the half-life of the substance.

For the New Horizons mission, we calculate how much Plutonium-238 is left after 9 years, between its launch in 2006 and its arrival at Pluto in 2015. Substituting the known values into the formula gives us an estimate of the remaining quantity of the isotope after this period.

Proportionality in Radioactive Decay

In radioactive decay, proportionality plays a crucial role in predicting power output, like in the case of the New Horizons spacecraft. The power output is directly proportional to the amount of radioactive material remaining. This means as Plutonium-238 decays over time, the power it produces decreases in a predictable manner.

To calculate the decrease in power output, we use the previously determined fraction of Plutonium-238 remaining. Since the original power output was 240 W at launch, the fraction of Plutonium-238 left after 9 years is used to find the new power output. The formula used is:\[ P(t) = P_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]where P_0 is the initial power and the remaining fraction is applied to calculate the power at the time of interest.

• This ensures that we can precisely determine the power available as the craft reaches its destination, which is essential for mission planning.