Understanding Acceleration Due to Gravity on a Planet

The question asks us to determine how the acceleration due to gravity on a planet compares to that on Earth, given that the planet has twice the mass and twice the radius of Earth. We are told that the acceleration due to gravity on the planet is $n$ times that on Earth, and we need to find the value of $n$.

Formula for Acceleration Due to Gravity

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

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

Where:

• \(G\) is the universal gravitational constant
• \(M\) is the mass of the planet
• \(R\) is the radius of the planet

Applying the Formula to Earth

Let's denote the mass of the Earth as \(M_E\) and the radius of the Earth as \(R_E\). The acceleration due to gravity on the surface of Earth (\(g_E\)) is:

\(g_E = \frac{GM_E}{R_E^2}\)

Applying the Formula to the Given Planet

Let's denote the mass of the planet as \(M_P\) and the radius of the planet as \(R_P\). According to the problem statement:

• The mass of the planet is twice the mass of Earth: \(M_P = 2M_E\)
• The radius of the planet is twice the radius of Earth: \(R_P = 2R_E\)

Now, let's calculate the acceleration due to gravity on the surface of this planet (\(g_P\)) using the formula:

\(g_P = \frac{GM_P}{R_P^2}\)

Substitute the given values for \(M_P\) and \(R_P\) in terms of \(M_E\) and \(R_E\):

\(g_P = \frac{G(2M_E)}{(2R_E)^2}\)

Now, let's simplify the expression:

\(g_P = \frac{G \cdot 2M_E}{4R_E^2}\)

\(g_P = \frac{2}{4} \cdot \frac{GM_E}{R_E^2}\)

\(g_P = \frac{1}{2} \cdot \frac{GM_E}{R_E^2}\)

Comparing Planet's Gravity to Earth's Gravity

We found that \(g_P = \frac{1}{2} \cdot \frac{GM_E}{R_E^2}\). From our earlier definition, we know that \(g_E = \frac{GM_E}{R_E^2}\). So, we can substitute \(g_E\) into the equation for \(g_P\):

\(g_P = \frac{1}{2} g_E\)

The question states that the acceleration due to gravity on the surface of the planet is \(n\) times that on Earth, which means \(g_P = n \cdot g_E\). By comparing this with our result \(g_P = \frac{1}{2} g_E\), we can see that the value of \(n\) is \(\frac{1}{2}\).

Conclusion on Gravity Calculation

The acceleration due to gravity on the surface of the described planet is half the acceleration due to gravity on the surface of Earth.

Thus, the value of \(n\) is \(\frac{1}{2}\).

Revision Table: Gravity Calculations

Additional Information: Factors Affecting Gravity

The acceleration due to gravity on the surface of a celestial body is directly dependent on its mass and inversely dependent on the square of its radius. This problem illustrates this relationship clearly. Let's consider some points:

• Mass: If the mass of a planet increases while its radius stays the same, the gravity on its surface increases proportionally.
• Radius: If the radius of a planet increases while its mass stays the same, the gravity on its surface decreases with the square of the radius. This means doubling the radius (with the same mass) reduces gravity to one-fourth.
• Combined Effect: In this problem, both mass and radius are doubled. The increased mass tends to increase gravity (by a factor of 2), while the increased radius tends to decrease gravity (by a factor of \(1/(2^2) = 1/4\)). The net effect is a change by a factor of \(2 \times \frac{1}{4} = \frac{1}{2}\).
• Newton's Law of Universal Gravitation: The formula for \(g\) is derived from Newton's Law of Universal Gravitation, which states that the force of attraction between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. For gravity on a planet's surface, one mass is the planet, the other is a small object on the surface, and the distance is the planet's radius. Force (\(F\)) = \(\frac{GMm}{R^2}\). Weight of an object (\(W\)) = \(mg\), where \(m\) is the object's mass. Since \(W = F\), \(mg = \frac{GMm}{R^2}\), which simplifies to \(g = \frac{GM}{R^2}\).