Compute the approximate mass of the Earth, assuming it to be...

Compute the approximate mass of the Earth, assuming it to be a sphere of radius $6370 \mathrm{~km}$. Ignore the planet's spin. Use $g=9.81$ $\mathrm{m} / \mathrm{s}^{2}$ and give your answer to three significant figures. Let $M_{E}$ be the mass of the Earth, and $m$ the mass of an object. The weight of the object on the planet's surface is equal to $m g .$ It is also equal to the gravitational force $c\left(m_{e} m\right), R_{i}^{2}$, where $R_{E}$ is the Earth's radius. Hence, $$ m g=G \frac{M_{E} m}{R_{E}^{2}} $$ from which $$ M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9,81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg} $$

Compute the approximate mass of the Earth, assuming it to be a sphere of radius $6370 \mathrm{~km}$. Ignore the planet's spin. Use $g=9.81$ $\mathrm{m} / \mathrm{s}^{2}$ and give your answer to three significant figures.

Let $M_{E}$ be the mass of the Earth, and $m$ the mass of an object. The weight of the object on the planet's surface is equal to $m g .$ It is also equal to the gravitational force $c\left(m_{e} m\right), R_{i}^{2}$, where $R_{E}$ is the Earth's radius. Hence,
$$
m g=G \frac{M_{E} m}{R_{E}^{2}}
$$
from which
$$
M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9,81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg}
$$

Key Concepts

Newton's Law of Universal Gravitation

This law states that every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. It provides the foundational equation used to relate gravitational forces to mass and distance, and it introduces the gravitational constant.

Gravitational Acceleration

Gravitational acceleration is the rate at which an object accelerates when falling solely under the influence of gravity, without other forces acting on it. It is directly linked to the gravitational force exerted by a massive body and serves as a measurable indicator of that body's gravitational strength at a given distance.

Relationship Between Weight and Gravitational Force

Weight is the force experienced by an object due to gravity, and it is calculated as the product of the object's mass and the gravitational acceleration. This relationship connects measurable weight to the underlying gravitational force described by Newton's law, enabling the calculation of a planet's mass when combined with knowledge of the distance from its center.

Spherical Symmetry in Gravitational Calculations

Assuming spherical symmetry simplifies the calculation of gravitational fields by allowing the mass of a large body to be treated as if it were concentrated at a point at its center. This approximation is fundamental when applying Newton's law of gravitation to celestial bodies, making it possible to relate surface measurements to overall mass.

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