So High Yet So Low

Classical Mechanics Level 3

A A and B B are points at a distance h h from the surface of the earth where the value of the acceleration due to gravity is the same. A A is above the surface and B B is below the surface.

If h = ( k − 1 2 ) R h=\left( \dfrac{ \sqrt{k}-1}{2}\right)R , where R R is the radius of the Earth, compute k k .


The answer is 5.

Shanthanu Rai by Shanthanu Rai , 21, India

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2 solutions

Magnitude of gravitational field under the surface of the earth: a = G M r R 3 a=\dfrac{GMr}{R^3} , and below the surface of the earth: a = G M r 2 a=\dfrac{GM}{r^2} , where r r is the distance from the center of the earth.

Then we have G M ( R − h ) R 3 = G M ( R + h ) 2 \dfrac{GM(R-h)}{R^3}=\dfrac{GM}{(R+h)^2} . On solving that we get:

( R − h ) ( R + h ) 2 = R 3 R 3 + R 2 h − R h 2 − h 3 = R 3 h 3 + R h 2 − R 2 h = 0 (R-h)(R+h)^2=R^3 \\ R^3+R^2h-Rh^2-h^3=R^3\\ h^3+Rh^2-R^2h=0

We discard the negative solutions, and using the quadratic formula we get h = 5 − 1 2 R h=\dfrac{\sqrt{5}-1}{2}R , hence k = 5 \boxed{k=5} .

5 Helpful 1 Interesting 1 Brilliant 0 Confused
Arjen Vreugdenhil
Apr 21, 2016

Such an elegant problem must have a golden solution... I guessed the 5 \sqrt 5 before doing the math. It's nice to see the golden ratio pop up-- again!

5 Helpful 1 Interesting 1 Brilliant 0 Confused

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