Question

. If the fastest passenger aircraft can fly 0.45 km/s (1000 mph), how long would it take to reach the sun? The galactic center?

Answer 1

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The passenger aircraft would take
`1.733* 10^(10)` years to reach the gallactic center.

Step-by-step explanation:

The distance to the sun from the Earth is approximately equal to
`1.496* 10^(8)\,km`, if the passenger travels at constant speed, then the time needed to reach the sun is calculated by the following kinematic formula:

`\Delta t = (s)/(v)` (1)

Where:

`s` - Travelled distance, measured in kilometers.

`v` - Speed of the passenger aircraft, measured in kilometers per second.

`\Delta t` - Travelling time, measured in seconds.

If we know that
`s = 1.496* 10^(8)\,km` and
`v = 0.45\,(km)/(s)`, then the travelling time is:

`\Delta t = (1.496* 10^(8)\,km)/(0.45\,(km)/(h) )`

`\Delta t = 3.324* 10^(8)\,s`

`\Delta t = 3847.736\,days`

`\Delta t = 10.542\,years`

The passenger aircraft would take 10.542 years to reach the Sun from the Earth.

The distance between the Earth and the galactic center is approximately equal to
`2.460* 10^(17)\,km`. If the passenger travels at constant speed and if we know that
`s = 2.460* 10^(17)\,km` and
`v = 0.45\,(km)/(s)` , then the travelling time is:

`\Delta t = (2.460* 10^(17)\,km)/(0.45\,(km)/(s) )`

`\Delta t = 5.467* 10^(17)\,s`

`\Delta t = 6.327* 10^(12)\,days`

`\Delta t = 1.733* 10^(10)\,years`

The passenger aircraft would take
`1.733* 10^(10)` years to reach the gallactic center.