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

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

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

= 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

= 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

something over 10 so ? /10

step-by-step explanation: