Understanding Acceleration Due to Gravity

The acceleration due to gravity on the surface of a planet depends on its mass and radius. It's a fundamental concept in physics, describing the force per unit mass experienced by an object on the planet's surface due to its gravitational pull.

The formula for the acceleration due to gravity ($g$) on the surface of a spherical body is given by:

$$g = \frac{GM}{R^2}$$

Where:

• \(G\) is the gravitational constant (approximately \(6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2\)).
• \(M\) is the mass of the planet.
• \(R\) is the radius of the planet.

Gravity on Earth vs. the Hypothetical Planet

Let's first consider Earth. We are given that the acceleration due to gravity on the surface of Earth is \(g\). Using the formula, we can write this as:

$$g_{\text{Earth}} = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} = g$$

Now, let's consider the hypothetical planet. We are told that its mass (\(M_{\text{Planet}}\)) and radius (\(R_{\text{Planet}}\)) are both half that of Earth. So:

• \(M_{\text{Planet}} = \frac{1}{2} M_{\text{Earth}}\)
• \(R_{\text{Planet}} = \frac{1}{2} R_{\text{Earth}}\)

The acceleration due to gravity on the surface of this hypothetical planet (\(g_{\text{Planet}}\)) can be calculated using the same formula:

$$g_{\text{Planet}} = \frac{GM_{\text{Planet}}}{R_{\text{Planet}}^2}$$

Substitute the given values for the planet's mass and radius in terms of Earth's mass and radius:

$$g_{\text{Planet}} = \frac{G \left(\frac{1}{2} M_{\text{Earth}}\right)}{\left(\frac{1}{2} R_{\text{Earth}}\right)^2}$$

Let's simplify the expression:

$$g_{\text{Planet}} = \frac{G \cdot \frac{1}{2} M_{\text{Earth}}}{\frac{1}{4} R_{\text{Earth}}^2}$$

We can rearrange the terms to separate the constants and the Earth terms:

$$g_{\text{Planet}} = \left(\frac{\frac{1}{2}}{\frac{1}{4}}\right) \left(\frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2}\right)$$

The fraction in the first parenthesis simplifies to:

$$\frac{\frac{1}{2}}{\frac{1}{4}} = \frac{1}{2} \times \frac{4}{1} = \frac{4}{2} = 2$$

The term in the second parenthesis is the acceleration due to gravity on Earth, which is given as \(g\):

$$\frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2} = g$$

Substitute these simplified values back into the expression for \(g_{\text{Planet}}\):

$$g_{\text{Planet}} = 2 \times g$$

So, the acceleration due to gravity on the hypothetical planet is \(2g\).

Step-by-Step Calculation Summary

Let \(M_E\) and \(R_E\) be Earth's mass and radius, and \(g_E = g\) be Earth's surface gravity.

Let \(M_P\) and \(R_P\) be the planet's mass and radius, and \(g_P\) be the planet's surface gravity.

1. Identify the formula for gravity: \(g = \frac{GM}{R^2}\).
2. Apply the formula to Earth: \(g_E = \frac{GM_E}{R_E^2} = g\).
3. Use the given relations for the planet: \(M_P = \frac{1}{2} M_E\) and \(R_P = \frac{1}{2} R_E\).
4. Apply the formula to the planet: \(g_P = \frac{GM_P}{R_P^2}\).
5. Substitute the planet's properties: \(g_P = \frac{G (\frac{1}{2} M_E)}{(\frac{1}{2} R_E)^2}\).
6. Simplify the expression: \(g_P = \frac{\frac{1}{2}}{\frac{1}{4}} \frac{G M_E}{R_E^2}\).
7. Calculate the numerical factor: \(\frac{\frac{1}{2}}{\frac{1}{4}} = 2\).
8. Substitute \(g = \frac{G M_E}{R_E^2}\) into the expression: \(g_P = 2 \cdot g\).
9. Final result: \(g_P = 2g\).

Comparing Gravity Values

We found that the acceleration due to gravity on the hypothetical planet is twice that on Earth. This happens because while the mass is halved (which would decrease gravity), the radius is also halved. Since gravity depends inversely on the square of the radius (\(1/R^2\)), halving the radius has a stronger effect (\(1/(1/2)^2 = 1/(1/4) = 4\)) than halving the mass (\(1/2\)). The combined effect is \(4 \times 1/2 = 2\).

Revision Table: Key Concepts in Gravity

Additional Information: Factors Affecting Gravity

The acceleration due to gravity on a planet's surface is primarily determined by its mass and radius, as shown by the formula \(g = \frac{GM}{R^2}\). However, other factors can cause slight variations in the measured value of \(g\) at different locations on the same planet:

• Altitude: Gravity decreases as you move away from the planet's surface (i.e., increase altitude). The formula becomes \(g(r) = \frac{GM}{r^2}\), where \(r\) is the distance from the planet's center (\(r = R + h\), where \(h\) is altitude).
• Shape of the planet: Planets are not perfect spheres. Earth, for example, is an oblate spheroid (bulges at the equator). This means points on the equator are slightly further from the center than points at the poles, resulting in slightly lower gravity at the equator.
• Rotation of the planet: The rotation of a planet causes a centrifugal effect, which slightly reduces the apparent gravity, especially near the equator. This effect is included when measuring the 'effective' \(g\).
• Local density variations: Variations in the density of the crust and mantle beneath the surface can cause minor local differences in gravity.

For idealized problems like the one discussed, we usually assume a perfectly spherical planet and consider gravity only on the surface, using the average radius and total mass.