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Igoryamba
3 years ago
8

. 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?

Geography
1 answer:
nadezda [96]3 years ago
3 0

Answer:

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

The passenger aircraft would take 1.733\times 10^{10} years to reach the gallactic center.

Explanation:

The distance to the sun from the Earth is approximately equal to 1.496\times 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 = \frac{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\times 10^{8}\,km and v = 0.45\,\frac{km}{s}, then the travelling time is:

\Delta t = \frac{1.496\times 10^{8}\,km}{0.45\,\frac{km}{h} }

\Delta t = 3.324\times 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\times 10^{17}\,km. If the passenger travels at constant speed and if we know that s = 2.460\times 10^{17}\,km and v = 0.45\,\frac{km}{s} , then the travelling time is:

\Delta t = \frac{2.460\times 10^{17}\,km}{0.45\,\frac{km}{s} }

\Delta t = 5.467\times 10^{17}\,s

\Delta t = 6.327\times 10^{12}\,days

\Delta t = 1.733\times 10^{10}\,years

The passenger aircraft would take 1.733\times 10^{10} years to reach the gallactic center.

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