Question

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested traveling at close to the speed of light. Suppose you wish to visit the red giant star Betelgeuse, which is$430ly$away, and that you want your $20,000kg$rocket to move so fast that you age only $20years$during the round trip.

a . How fast, as a fraction of c, must the rocket travel relative to earth?

b. How much energy is needed to accelerate the rocket to this speed?

c. Compare this amount of energy to the total energy used by the United States in the year $2015$, which was roughly$1.0\times {10}^{20}J$

Short answer

a. The speed is $0.998C$.

b. The required energy is $7.34\times {10}^{22}J$.

c. The amount of energy is$734$times more than the United States in the year$2015$.

Step-by-step solution

Part (a) step 1: Given Information 

We need to find speed, as a fraction of c, must the rocket travel relative to the earth .

Part (a) step 2:Simplify

Suppose the voyage is moving at the speed $v$, now since the speed is very close to $C$, the time taken from the earth's frame of reference should be $430y$. But in voyage's frame of reference, it is $20y$.So $\gamma$is given as $\lambda =\frac{2\times 430}{20}=43$Now we know that

$$\begin{gathered} \lambda =\frac{1}{\sqrt{1-{\beta }^2}} \\ \Rightarrow 43=\frac{1}{\sqrt{1-{\beta }^2}} \\ \Rightarrow \beta =0.9997 \end{gathered}$$

Now we know that

$$\begin{gathered} \beta =\frac{\nu }{C} \\ \Rightarrow v=\beta C=0.9997C \end{gathered}$$

Part (b) step 2: Given Information 

We need to find how much energy is needed to accelerate the rocket to this speed.

Part (b) step 2: Simplify

The kinetic energy of the rocket is given by$E_k=\frac{m{v}^2}{\sqrt{1-{\beta }^2}}=\frac{2000\times {\left(0.9997\times 3\times {10}^8\right)}^2}{\sqrt{1-0.{9997}^2}}=7.34\times {10}^{22}J$

Part (c) step 3: Given Information 

We need to Compare this amount of energy to the total energy used by the United States in the year $2015$, which was roughly $1.0\times {10}^{20}J$.

Part (c) step 3: Simplify

The energy is $\frac{7.34\times {10}^{22}}{1.0\times {10}^{20}}=734$times more than the annual consumption of energy in the year$2010$.