Question

A piano is hauled into space a distance 3 earth radii above the surface of the north pole and is then dropped. About how long will it take to splash down in the arctic ocean

Answer 1

Answer:

The answer to the question is

It would take about 167.021 s  to splash down in the arctic ocean.

Step-by-step explanation:

The gravitational force is given by

$F_G = \frac{G*m_1*m_2}{r^2}$

Where m₁ mass of the piano and

m₂ = mass of the Earth

r = 3·R where R = radius of the earth

as stated in the question we have varying acceleration due to the inverse square law

Therefore at 3 × Radius of the earth we have

$F_G$ = 109.083 N and the acceleration =1.09083 m/s²

If the body falls from 3·R to 2·R with that acceleration we have

S = u·t +0.5×a·t²  = 0.5×a·t² as u = 0

That is 6371 km = 0.5·1.09083·t²

t₁ = 108.079 s and we have

v₁² = u₁² +2·a₁·s₁ =  2·a₁·s₁ = 117.895 m/s

For the next stage r₂ = 2R

Therefore F = 245.436 N and a₂ = F/m₁ = 2.45436 m/s²

Therefore the time from 2R to R is given by

S₂ =R=u·t+0.5·a₂·t² = v₁·t + 0.5·a₂·t²

or 6371 km = 117.895 m/s × t + 0.5 × 2.45436 × t²

Which gives 1.22718 × t² + 117.895 × t -6371  = 0

Factorizing we have (t+134.631)(t-38.56)×1.22718 = 0

Therefore t = -134.631 s or 38.56 s as we only deal with positive values of time in the present question we have t₂ = 38.56 s and

v₂² = v₁² + 2·a₂·S  = (117.895 m/s)² + 2·2.45436 m/s²×6371 km = 45172.686

v₂ = 212.54 m/s

For final stage we have r = R and

$F_{G3}$ = 981.746 N and a₂ = F/m₁ = 9.81746 m/s²

Therefore the time from R to the arctic ocean  is given by

S₃ =R=v₂·t+0.5·a₂·t² = 212.54·t + 0.5·9.81746·t² = 6371

Which gives

Therefore t₃ = 20.382 s

Therefore, it will take about t₁ + t₂ + t₃ = 108.079 s + 38.56 s + 20.382 s  = 167.021 s  to splash down in the ocean