Question

A famous boxer has a mass of M=71.1 kg and needs to weigh in for a fight to make sure that he still qualifies for a particular weight class. He knows, however, that he still needs to lose m=4.74 g of mass (not weight!) in order to qualify. Luckily for him, he knows physics. This boxer knows that he can use his fame to influence where the weigh-in (which will be televised, so he doesn't want to embarrass himself) will take place. He knows that the scale used for these weigh-ins is calibrated to the gravitational field at sea-level. How high (in meters) above sea-level should he petition the weigh-in to take place so that he will just barely qualify for his desired weight class?

Answer 1

Answer:

333 m

Explanation:

Gravitational force is:

F = GMm / r²,

where G is the universal gravitational constant,

M is the mass of the Earth,

m is the mass of the object,

and r is the distance between the object and the Earth's center.

We want the weight of the boxer at the new elevation (r+h) to be equal to the weight of a lighter boxer at sea level.

F₁ = F₂

GMm₁ / r² = GMm₂ / (r+h)²

m₁ / r² = m₂ / (r+h)²

(r+h)² = m₂ r² / m₁

r + h = r √(m₂ / m₁)

h = -r + r √(m₂ / m₁)

h = r (-1 + √(m₂ / m₁))

Given that the radius of the Earth is r = 6.37×10⁶ m, the mass of the boxer is m₂ = 71.1 kg, and the desired mass is m₁ = 71.1 – 0.00474 = 71.09526 kg:

h = (6.37×10⁶) (-1 + √(71.1 / 71.09526))

h = 333 m

The boxer would have to choose a place 333 m above sea level to appear 4.74 g lighter.