Question

Be sure to show all calculations clearly and state your final answers in complete sentences. Long Trips at Constant Acceleration. Consider a spaceship on a long trip with a constant acceleration of \(1 g\). Although the derivation is beyond the scope of this book, it is possible to show that, as long as the ship is gone from Earth for many years, the amount of time that passes on the spaceship during the trip is approximately \\[ t_{\text {ship }}=\frac{2 c}{g} \ln \left(\frac{g \times D}{c^{2}}\right) \\] where \(D\) is the distance to the destination and \(\ln\) is the natural logarithm. If \(D\) is in meters, \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and \(c=3 \times 10^{8} \mathrm{m} / \mathrm{s}\) the answer will be in units of seconds. Use this formula as needed to answer the following questions. Be sure to convert the distances from light-years to meters and final answers from seconds to years; useful conversions: 1 light-year \(\approx 9.5 \times 10^{15} \mathrm{m}\) \(1 \mathrm{yr} \approx 3.15 \times 10^{7} \mathrm{s}\). a. Suppose the ship travels to a star that is 500 light-years away. How much time will pass on the ship? Approximately how much time will pass on Earth? Explain. b. Suppose the ship travels to the center of the Milky Way Galaxy, about 28,000 light-years away. How much time will pass on the ship? Compare this to the amount of time that passes on Earth. c. The Andromeda Galaxy is about 2.2 million light-years away. Suppose you had a spaceship that could constantly accelerate at \(1 g .\) Could you go to the Andromeda Galaxy and back within your lifetime? Explain. If you could make the journey, what would you find when you returned to Earth?

Short answer

Time on the ship: (a) ~12.1 years, (b) ~93.6 years, (c) ~105.4 years. Earth time: (a) 1000 years, (b) 56,000 years, (c) 4,400,000 years. The journey to Andromeda and back cannot be done within a human lifetime when considering Earth time.

Step-by-step solution

Convert Distance to Meters for Part (a)

The star is 500 light-years away. Use the conversion factor: 1 light-year is approximately \(9.5 \times 10^{15}\) meters. So, the distance \(D\) in meters is:\[D = 500 \times 9.5 \times 10^{15} = 4.75 \times 10^{18} \text{ meters}\]

Calculate Time on Ship for Part (a)

Using the formula given:\[t_{\text{ship}} = \frac{2c}{g} \ln\left(\frac{g \times D}{c^{2}}\right)\]Substitute \(g = 9.8 \, \text{m/s}^2\), \(c = 3 \times 10^{8} \, \text{m/s}\), and \(D = 4.75 \times 10^{18} \, \text{m}\):\[t_{\text{ship}} = \frac{2 \times 3 \times 10^{8}}{9.8} \ln\left(\frac{9.8 \times 4.75 \times 10^{18}}{(3 \times 10^{8})^2}\right)\]This simplifies to:\[t_{\text{ship}} \approx \frac{6 \times 10^{8}}{9.8} \ln\left(\frac{4.655 \times 10^{19}}{9 \times 10^{16}}\right)\]\[t_{\text{ship}} \approx 6.12 \times 10^7 \ln\left(516.11\right)\]\[t_{\text{ship}} \approx 6.12 \times 10^7 \times 6.25 \approx 3.825 \times 10^8 \, \text{seconds}\]

Convert Ship Time to Years for Part (a)

Convert the time from seconds to years using the conversion factor: 1 year is approximately \(3.15 \times 10^7\) seconds.\[t_{\text{ship}} \approx \frac{3.825 \times 10^8}{3.15 \times 10^7} \approx 12.14 \, \text{years}\]

Calculate Earth Time for Part (a)

The Earth time will be approximately twice the light-travel time since the ship accelerates to the halfway point and then decelerates:\[t_{\text{earth}} = 2 \times \text{Distance in light-years} = 2 \times 500 = 1000 \text{ years}\]

Repeat Steps 1 to 3 for Part (b)

Convert \(D = 28,000 \text{ light-years}\) to meters:\[D = 28,000 \times 9.5 \times 10^{15} = 2.66 \times 10^{20} \text{ meters}\]Calculate \(t_{\text{ship}}\):\[t_{\text{ship}} \approx \frac{6 \times 10^{8}}{9.8} \ln\left(\frac{9.8 \times 2.66 \times 10^{20}}{(3 \times 10^{8})^2}\right)\]This simplifies to:\[t_{\text{ship}} \approx 6.12 \times 10^7 \ln\left(2.92 \times 10^{21}\right)\]\[t_{\text{ship}} \approx 6.12 \times 10^7 \times 48.29 \approx 2.95 \times 10^9 \, \text{seconds}\]Convert to years:\[t_{\text{ship}} \approx \frac{2.95 \times 10^9}{3.15 \times 10^7} \approx 93.6 \, \text{years}\]

Calculate Earth Time for Part (b)

The Earth time is:\[t_{\text{earth}} = 2 \times 28,000 = 56,000 \text{ years}\]

Repeat Steps 1 to 3 for Part (c)

Convert \(D = 2.2 \, \text{million light-years}\) to meters:\[D = 2.2 \times 10^{6} \times 9.5 \times 10^{15} = 2.09 \times 10^{22} \text{ meters}\]Calculate \(t_{\text{ship}}\):\[t_{\text{ship}} \approx \frac{6 \times 10^{8}}{9.8} \ln\left(\frac{9.8 \times 2.09 \times 10^{22}}{(3 \times 10^{8})^2}\right)\]This simplifies to:\[t_{\text{ship}} \approx 6.12 \times 10^7 \ln\left(7.265 \times 10^{23}\right)\]\[t_{\text{ship}} \approx 6.12 \times 10^7 \times 54.18 \approx 3.32 \times 10^9 \, \text{seconds}\]Convert to years:\[t_{\text{ship}} \approx \frac{3.32 \times 10^9}{3.15 \times 10^7} \approx 105.4 \, \text{years}\]

Calculate Earth Time for Part (c)

The Earth time is twice the distance in light-years, as the journey is a round trip:\[t_{\text{earth}} = 2 \times 2.2 \times 10^{6} = 4.4 \times 10^{6} = 4,400,000 \text{ years}\]

Key concepts

constant acceleration

In the context of spaceship physics, constant acceleration is vital when exploring long-distance space travel. It refers to a scenario where the rate of change of velocity (acceleration) remains the same throughout the journey. This is particularly interesting in the scope of theoretical interstellar travel, where a spaceship might continuously accelerate at a value equivalent to Earth's gravitational acceleration, known as \(1g\) or \(9.8 \, \text{m/s}^2\).
This constant acceleration provides a convenient way of maintaining passengers' comfort by simulating Earth-like gravity during long trips. By consistently accelerating to the halfway point and decelerating the rest of the way, the speed and overall time taken to reach distant celestial bodies can be considerably lessened compared to uniform velocity travel. This concept allows us to employ natural logarithms and conversion factors to calculate time experienced by travelers versus time on Earth.

relativity

Relativity, particularly Einstein’s theory of special relativity, plays a key role in understanding time experienced differently by travelers aboard a spaceship compared to observers on Earth. Two fundamental insights from relativity are crucial here:

• Time Dilation: As objects approach the speed of light, time for those objects appears to slow compared to observers at rest. This means astronauts would age more slowly than people remaining on Earth.
• Simultaneity: Events observed from different frames of reference may not occur in the same order. This impacts how time is perceived for travelers versus those at rest on Earth.

For practical calculations, relativity ensures that the spaceship's journey time is calculated using the formula encapsulated for constant acceleration. The foundational theory behind such equations stems from understanding how relative velocities and distances affect time perception.

light-years to meters conversion

When planning interstellar travel in spaceship physics, distances are often given in light-years because they encompass vast scales that are impractical to express in smaller units. A light-year signifies the distance that light travels in one year, approximately \(9.5 \times 10^{15}\) meters.
Converting light-years to meters becomes essential when applying formulas that require meter as the distance unit, as is the case in our exercise. The conversion process involves simple multiplication:

• First, denote the number of light-years as a numeral.
• Multiply by the conversion factor \(9.5 \times 10^{15}\) to obtain meters.

This conversion enables accurate application of physics equations which then produce usable results for spacecraft distances, rendering complex, astronomical measurements into manageable numbers for calculations.

natural logarithm calculation

The natural logarithm, denoted as \(\ln\), is a crucial mathematical tool used in physics to linearize exponential processes. In the formula for time experience by a spaceship with constant acceleration, the natural logarithm helps compute the time difference due to varying velocity and distance.
Specifically, we use \(\ln\) in the following context of our equation:

• The argument of the natural logarithm relates to the ratio of the spaceship’s trip influenced by gravitational forces and light speed limitations.
• The size of this calculation directly affects the resulting time perceived by travelers aboard.

Calculating the natural logarithm involves using a calculator or software that applies Euler’s number \(e\), where \(\ln(x)\) equals the power to which \(e\) must be raised to obtain \(x\). Emphasizing its application in these equations transforms complex multiplicative and exponential relationships into linear ones, simplifying subsequent calculations in space travel scenarios.