Answer:
The passenger aircraft would take 10.542 years to reach the Sun from the Earth.
The passenger aircraft would take [tex]1.733\times 10^{10}[/tex] years to reach the gallactic center.
Explanation:
The distance to the sun from the Earth is approximately equal to [tex]1.496\times 10^{8}\,km[/tex], if the passenger travels at constant speed, then the time needed to reach the sun is calculated by the following kinematic formula:
[tex]\Delta t = \frac{s}{v}[/tex] (1)
Where:
[tex]s[/tex] - Travelled distance, measured in kilometers.
[tex]v[/tex] - Speed of the passenger aircraft, measured in kilometers per second.
[tex]\Delta t[/tex] - Travelling time, measured in seconds.
If we know that [tex]s = 1.496\times 10^{8}\,km[/tex] and [tex]v = 0.45\,\frac{km}{s}[/tex], then the travelling time is:
[tex]\Delta t = \frac{1.496\times 10^{8}\,km}{0.45\,\frac{km}{h} }[/tex]
[tex]\Delta t = 3.324\times 10^{8}\,s[/tex]
[tex]\Delta t = 3847.736\,days[/tex]
[tex]\Delta t = 10.542\,years[/tex]
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 [tex]2.460\times 10^{17}\,km[/tex]. If the passenger travels at constant speed and if we know that [tex]s = 2.460\times 10^{17}\,km[/tex] and [tex]v = 0.45\,\frac{km}{s}[/tex] , then the travelling time is:
[tex]\Delta t = \frac{2.460\times 10^{17}\,km}{0.45\,\frac{km}{s} }[/tex]
[tex]\Delta t = 5.467\times 10^{17}\,s[/tex]
[tex]\Delta t = 6.327\times 10^{12}\,days[/tex]
[tex]\Delta t = 1.733\times 10^{10}\,years[/tex]
The passenger aircraft would take [tex]1.733\times 10^{10}[/tex] years to reach the gallactic center.
