To what fraction of its current radius would Earth have to shrink (with no change in mass) for the gravitational acceleration at its surface to triple?

Gravitational force is an invisible pull that draws objects towards each other. On Earth, this is the force that keeps us rooted to the ground and causes objects to fall when dropped. This force can be explained by Isaac Newton's universal law of gravitation, which states that every mass exerts an attractive force on every other mass.

The formula to calculate this force is
\[ F = G \cdot \frac{m_1 \cdot m_2}{r^2} \]
where:

• \(F\) is the gravitational force between two masses,
• \(G\) is the gravitational constant (\(6.674 \times 10^{-11} N \cdot (m/kg)^2\)),
• \(m_1\) and \(m_2\) are the masses of the two objects, and
• \(r\) is the distance between the centers of the two masses.

When discussing gravitational acceleration, such as on Earth's surface, the term \(g\) is commonly used, and it represents the acceleration due to gravity that any object feels when in the vicinity of a massive body like the Earth. In the exercise to triple the gravitational acceleration, we don't change the mass of Earth but alter its radius, showing how deeply interconnected gravitational force and distance are.

The relationship between mass and radius is crucial to understanding gravitational forces. The gravitational acceleration at the surface of a planet is directly influenced by both its mass and radius, as shown in the formula for gravitational force. The formula for gravitational acceleration is:
\[ g = G \cdot \frac{M}{r^2} \]
Here, \(g\) is the gravitational acceleration, \(M\) is the planet's mass, \(G\) is the universal gravitational constant, and \(r\) is the radius from the center of the mass to the point where \(g\) is being measured.

If a planet's radius is decreased while its mass remains constant, the gravitational acceleration at its surface increases. This happens because as the radius \(r\) decreases, the value of \(\frac{1}{r^2}\) increases, thereby increasing \(g\). In the exercise, we needed to find what fraction of Earth's current radius would result in the gravitational acceleration tripling, indicating a significant inverse square relationship between radius and gravitational acceleration.

Newton's law of gravitation is a fundamental principle that describes the gravitational attraction between objects with mass. According to Newton, every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
This is written as:
\[F = G \cdot \frac{m_1 \cdot m_2}{r^2}\]
where \(F\) represents the force of attraction, \(m_1\) and \(m_2\) are the masses, \(r\) is the distance between the centers of the masses, and \(G\) is the gravitational constant.

This law explains not only the force between two objects but also how the gravitational force influences the motion of celestial bodies and the trajectories of spacecraft. It's the foundational concept necessary for calculating orbital mechanics and understanding the force that keeps planets in orbit around stars. The exercise we looked at is an application of this law, where we examine the effect of changing the radius on the gravitational acceleration on Earth's surface -- a practical demonstration of how altering distance impacts the force of gravity in line with Newton's law.