Question

A 100 kg scientist finds that the acceleration of gravity at the north and south pole is measured to be 9.832 m/s^2 and only 9.780 m/s^2 at the equator. If the mass of the Earth is 5.98x10^24 kg, determine the radius of the Earth at the equator and at the poles.

Answer 1

Answer on Question 49758, Physics, Mechanics | Kinematics | Dynamics

Question:

A $100kg$ scientist finds that the acceleration of gravity at the north and south pole is measured to be $9.832\frac{m}{s^2}$ and only $9.780\frac{m}{s^2}$ at the equator. If the mass of the Earth is $5.98 \cdot 10^{24}kg$, determine the radius of the Earth at the equator and at the poles.

Solution:

By the definition of the acceleration of gravity we have:

$$g = G \frac {M _ {E}}{R ^ {2}},$$

where $g$ is the acceleration of gravity, $G = 6.67 \cdot 10^{-11} \frac{Nm^2}{kg^2}$ is the gravitational constant, $M_E = 5.98 \cdot 10^{24} kg$ is the mass of the Earth and $R$ is the radius of the Earth.

From this formula we can find the radius of the Earth at the equator and at the poles:

$$R _ {eq} = \sqrt {G \frac {M _ {E}}{g _ {eq}}} = \sqrt {\frac {6 . 6 7 \cdot 1 0 ^ {- 1 1} \frac {N m ^ {2}}{k g ^ {2}} \cdot 5 . 9 8 \cdot 1 0 ^ {2 4} k g}{9 . 7 8 0 \frac {m}{s ^ {2}}}} = 6. 3 8 6 2 2 3 \cdot 1 0 ^ {6} m.$$ $$R _ {pole} = \sqrt {G \frac {M _ {E}}{g _ {pole}}} = \sqrt {\frac {6 . 6 7 \cdot 1 0 ^ {- 1 1} \frac {N m ^ {2}}{k g ^ {2}} \cdot 5 . 9 8 \cdot 1 0 ^ {2 4} k g}{9 . 8 3 2 \frac {m}{s ^ {2}}}} = 6. 3 6 9 3 1 2 \cdot 1 0 ^ {6} m.$$

Answer:

a) $R_{eq} = 6.386223 \cdot 10^{6} m.$

b) $R_{pole} = 6.369312 \cdot 10^{6} m.$

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