Mass from weight?

Classical Mechanics Level pending

Allan is standing on a perfectly spherical asteroid. He decides to perform an experiment to calculate it's mass.

He measures the gravity of the asteroid at different heights and determines that if he were to jump 100 metres upwards his weight would be $\frac{7}{8}$ of what it was on the surface of the asteroid.

Given that the asteroid is made entirely out of iron, find the mass of the asteroid in Kilograms.

Details and assumptions :

Take the density of iron to be $7850 \text{ kgm}^{-3}$ .
Give your answer to five significant figures in scientific notation. E.g. if your got $123456789$ you would answer with $1.2345\text{E+8}$
Assume Allan's centre of mass is on the asteroids surface.

The answer is 9.9899E+13.

1 solution

Jack Rawlin
Feb 22, 2016

Let $F$ be the weight of Allan while on the surface of the asteroid.

$F = \frac{Gm_1m_2}{r^2}$

$\frac{7}{8}F = \frac{Gm_1m_2}{(r + 100)^2}$

$\frac{7}{8}\left(\frac{Gm_1m_2}{r^2}\right) = \frac{Gm_1m_2}{(r + 100)^2}$

$\frac{7}{8r^2} = \frac{1}{(r + 100)^2}$

$\frac{8r^2}{7} = (r + 100)^2$

$\frac{8r^2}{7} = r^2 + 200r + 10000$

$8r^2 = 7r^2 + 1400r + 70000$

$r^2 - 1400r - 70000 = 0$

$r = \frac{1400 \pm \sqrt{(-1400)^2 - 4(1)(-70000)}}{2}$

$r = 1448.3314\cdots,~ -48.3314\cdots$

$r > 0 \therefore r \neq -48.3314\cdots$

$V = \frac{4}{3}\pi r^3$

$V = \frac{4}{3}\pi \cdot (1448.3314\cdots)^3$

$V = (1.2726\cdots) \cdot 10^{10}$

$m = V\rho$

$m = ((1.2726\cdots) \cdot 10^{10}) \cdot 7850$

$m = (9.9899\cdots) \cdot 10^{13}$

$\large \boxed{m = 9.9899\text{E+13}}$

Mass from weight?

The answer is 9.9899E+13.

1 solution

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