Question

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v = 239 m/s. Upon landing, the plane can produce an average deceleration of a = 16.5 m/s².
How long will it take the plane to circle the Earth at the equator?

Answer 1

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

$R= 6370*10^3 m$

$v = 239m/s$

$a = 16.5m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$, then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}$

$t = 167463.97s$

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds