Understanding Gravity on Planets with Same Density

This question asks us to compare the acceleration due to gravity on the surface of two planets that have the same density but different radii. We are given that Planet 1 has radius \(R_1\) and Planet 2 has radius \(R_2\), and \(R_1 > R_2\). The density of both planets is the same, let's call it \(\rho\).

Calculating Acceleration Due to Gravity

The acceleration due to gravity (\(g\)) on the surface of a planet with mass \(M\) and radius \(R\) is given by Newton's law of gravitation:

\(g = \frac{GM}{R^2}\)

where \(G\) is the universal gravitational constant.

Relating Mass to Density and Radius

Since we are given the density, we need to express the mass \(M\) in terms of density (\(\rho\)) and volume. Assuming the planets are perfect spheres, the volume \(V\) of a planet with radius \(R\) is given by:

\(V = \frac{4}{3}\pi R^3\)

The mass \(M\) of the planet is the product of its density and volume:

\(M = \rho V = \rho \left(\frac{4}{3}\pi R^3\right)\)

Deriving Gravity in Terms of Density and Radius

Now, we can substitute the expression for mass (\(M\)) into the formula for acceleration due to gravity (\(g\)):

\(g = \frac{G \left(\rho \frac{4}{3}\pi R^3\right)}{R^2}\)

Simplify the expression:

\(g = G \rho \frac{4}{3}\pi \frac{R^3}{R^2}\)

\(g = \frac{4}{3}\pi G \rho R\)

Comparing Gravity for the Two Planets

Using this derived formula, we can write the acceleration due to gravity for Planet 1 (\(g_1\)) and Planet 2 (\(g_2\)). Since both planets have the same density (\(\rho\)), we have:

• For Planet 1: \(g_1 = \frac{4}{3}\pi G \rho R_1\)
• For Planet 2: \(g_2 = \frac{4}{3}\pi G \rho R_2\)

We are given that \(R_1 > R_2\). Let's compare \(g_1\) and \(g_2\):

\(\frac{g_1}{g_2} = \frac{\frac{4}{3}\pi G \rho R_1}{\frac{4}{3}\pi G \rho R_2}\)

Since \(\frac{4}{3}\pi G \rho\) is a common constant for both planets, it cancels out:

\(\frac{g_1}{g_2} = \frac{R_1}{R_2}\)

As \(R_1 > R_2\), this means \(\frac{R_1}{R_2} > 1\).

Therefore, \(\frac{g_1}{g_2} > 1\), which implies \(g_1 > g_2\).

Conclusion on Acceleration Due to Gravity

When two planets have the same density, the acceleration due to gravity on their surface is directly proportional to their radius. Since Planet 1 has a larger radius than Planet 2, the acceleration due to gravity on the surface of Planet 1 is greater than that on Planet 2.

The relationship is \(g_1 > g_2\).

Revision Table: Gravity and Planetary Properties

Additional Information: Factors Affecting Surface Gravity

The acceleration due to gravity on the surface of a planet depends on its mass and radius. However, as shown in this problem, it can also be related to its density and radius. Here's a bit more detail:

• Mass (M) and Radius (R): The standard formula \(g = \frac{GM}{R^2}\) shows that gravity increases with mass and decreases with the square of the radius. A massive planet can have high gravity, but if it's very large, the gravity on its surface (far from the center) might be less than expected.
• Density (\(\rho\)) and Radius (R): The formula \(g = \frac{4}{3}\pi G \rho R\) is particularly useful when comparing planets of similar composition (hence similar density). It highlights that for planets of the same density, gravity is simply proportional to how big they are. A larger radius means a greater distance over which mass is distributed, but the effect of the increasing mass (as \(R^3\)) outweighs the increasing distance squared (as \(R^2\)), resulting in a linear dependence on \(R\).
• Other Factors: Real-world surface gravity can also be affected by the planet's rotation (centrifugal force reduces apparent gravity at the equator), local geological variations (like mountains or dense ore deposits), and altitude above the reference surface.