Question

Astronomers discover a planet orbiting around a star similar to our sun that is 35 light years away. How fast must a rocket ship go if the round trip is to take no longer than 70 years in time for the astronauts aboard

Answer 1

To make a 70-year round trip to a planet 35 light-years away, a spaceship must travel at a speed of about 1 light-year per year, which is equivalent to 9,460,800,000,000 miles per year. This calculation presumes relativistic effects such as time dilation are negligible for the astronauts.

Step-by-step explanation:

The question involves calculating the required speed for a round trip to a planet orbiting a star 35 light-years away in no more than 70 years from the perspective of the astronauts. To achieve this, the spacecraft must travel at a speed that allows the round trip distance of 70 light-years to be covered in 70 years. Given that light travels at 186,000 miles per second, and there are various units of time that factor into the calculation of a light-year (60 seconds in a minute, 60 minutes in an hour, 24 hours in a day, and 365 days in a year), we can calculate the required speed as follows:

Speed = Total Distance / Total Time

Speed = 70 light-years / 70 years

To find the speed in miles per year, we first find the distance per light-year:

Distance per light-year = 186,000 miles/second * 60 seconds/minute * 60 minutes/hour * 24 hours/day * 365 days/year

The speed, therefore, must be this distance per light year (9,460,800,000,000 miles per light-year) divided by the number of years.

Moreover, the closer a spaceship travels to the speed of light, the more significant the effects of time dilation as per Einstein's theory of relativity, which means the astronauts would age less than the time that passes on Earth.

Answer 2

`v = 0.7071c`

Step-by-step explanation:

Given

Distance to the planet = 35 light years. So, the entire distance is: 2 * 35 = 70.

`\triangle{x'} = 70`

`T_0 = 70\ years` i.e time of travel of the ship.

For the observer on earth, the time is:

`T' = \gamma T_0`

The required speed so that it does not take more than 70 years is then calculated using:

`\triangle x' = vT'`

Substitute
`T' = \gamma T_0`

`\triangle x' = v\gamma T_0`

`\gamma = (1)/(√(1 - v^2/c^2))`

So, we have:

`\triangle x' = (vT_0)/(√(1 - v^2/c^2))`

Make v the subject of formula.

Square both sides

`\triangle x'^2 = (v^2T^2_0)/(1 - v^2/c^2)`

Cross Multiply

`(1 - (v^2)/(c^2)) *\triangle x'^2 = v^2T^2_0`

Divide both sides by
`\triangle x'^2`

`(1 - (v^2)/(c^2)) = (v^2T^2_0)/(\triangle x'^2)`

Divide through by
`v^2`

`((1)/(v^2) - (v^2)/(v^2*c^2)) = (v^2T^2_0)/(v^2\triangle x'^2)`

`(1)/(v^2) - (1)/(c^2) = (T^2_0)/(\triangle x'^2)`

Make
`(1)/(v^2)` the subject

`(1)/(v^2) = (T^2_0)/(\triangle x'^2) + (1)/(c^2)`

Inverse both sides

`v^2 = (1)/((T^2_0)/(\triangle x'^2) + (1)/(c^2))`

Take square root of both sides

`v = \sqrt{(1)/((T^2_0)/(\triangle x'^2) + (1)/(c^2))}`

Substitute values for
`T_0` and
`\triangle x`

`v = \sqrt{(1)/((70^2)/((70c)^2) + (1)/(c^2))}`

`v = \sqrt{(1)/((70^2)/(70^2*c^2) + (1)/(c^2))}`

`v = \sqrt{(1)/((1)/(c^2) + (1)/(c^2))}`

`v = \sqrt{(1)/((2)/(c^2))}`

`v = \sqrt{(c^2)/(2)}`

`v = c\sqrt{(1)/(2)}`

`v = c * 0.7071`

`v = 0.7071c`