Understanding Acceleration Due to Gravity

Acceleration due to gravity is the acceleration experienced by an object due to the gravitational force exerted by a celestial body, such as the Earth. At the surface of the Earth, this acceleration is commonly denoted by 'g'.

Deriving the Formula for Acceleration Due to Gravity

According to Newton's Law of Universal Gravitation, the force of attraction (F) between two objects of masses \(m_1\) and \(m_2\) separated by a distance \(r\) is given by:

\(F = \frac{G m_1 m_2}{r^2}\)

where G is the universal gravitational constant.

Consider an object of mass 'm' placed on the surface of the Earth. The Earth has mass M and radius R. The distance between the center of the Earth and the object on its surface is approximately equal to the Earth's radius, R.

The gravitational force exerted by the Earth on the object is:

\(F = \frac{G M m}{R^2}\)

According to Newton's second law of motion, the force F acting on the object is also given by \(F = ma\), where 'a' is the acceleration of the object. In this case, the acceleration 'a' is the acceleration due to gravity 'g'.

So, we have:

\(F = mg\)

Equating the two expressions for force:

\(mg = \frac{G M m}{R^2}\)

We can cancel out the mass of the object 'm' from both sides:

\(g = \frac{G M}{R^2}\)

Proportionality of Acceleration Due to Gravity

The formula for acceleration due to gravity at the surface of the Earth is \(g = \frac{G M}{R^2}\). In this formula:

• G is the universal gravitational constant, which is a constant value.
• M is the mass of the Earth.
• R is the radius of the Earth.

From the formula, we can see how 'g' depends on M and R. Since G is constant, 'g' is directly proportional to the mass of the Earth (M) and inversely proportional to the square of the radius of the Earth (R). We can express this proportionality as:

\(g \propto \frac{M}{R^2}\)

This means if the mass M increases, g increases, and if the radius R increases, g decreases (because it's in the denominator squared). Therefore, the acceleration due to gravity at the surface of the Earth is proportional to \( \frac{M}{R^2} \).

Comparing with Options

Let's look at the given options and compare them with our derived proportionality:

1. \( \frac{M}{R} \) - This is not proportional to g.
2. MR - This is not proportional to g.
3. \( \frac{M}{R^2} \) - This matches our derived proportionality \( g \propto \frac{M}{R^2} \).
4. \( \frac{M^2}{R} \) - This is not proportional to g.

Thus, the acceleration due to gravity at the surface of the earth is proportional to \( \frac{M}{R^2} \).

Revision Table: Key Concepts of Acceleration Due to Gravity

Additional Information: Variations in Acceleration Due to Gravity

While the formula \( g = \frac{GM}{R^2} \) gives the acceleration due to gravity at the surface, the value of 'g' is not constant everywhere. It varies due to several factors:

• Altitude: As you go above the Earth's surface, the distance from the center increases (R + h). The acceleration due to gravity decreases with altitude.
• Depth: As you go below the Earth's surface, the mass contributing to gravity decreases, and the distance from the center changes. The acceleration due to gravity also decreases with depth (assuming uniform density).
• Shape of Earth: The Earth is not a perfect sphere; it is an oblate spheroid (bulges at the equator). The radius is larger at the equator than at the poles. Since \( g \propto \frac{1}{R^2} \), 'g' is slightly less at the equator and more at the poles.
• Rotation of Earth: The rotation of the Earth causes a centrifugal force, which reduces the apparent weight of objects and thus the effective acceleration due to gravity, especially at the equator.

The question specifically asks about the proportionality at the surface of the Earth, for which the formula \( g = \frac{GM}{R^2} \) is applicable, leading to the proportionality \( g \propto \frac{M}{R^2} \).