If the acceleration due to gravity on the surface of earth is g, then the acceleration due to gravity on the surface of a planet whose mass is same as that of earth and radius is twice as that of earth is ___________.
g/4
Calculating Acceleration Due to Gravity on a Planet
The question asks us to find the acceleration due to gravity on the surface of a planet given its mass and radius relative to Earth. We are told that the planet has the same mass as Earth but twice the radius. The acceleration due to gravity on the surface of Earth is given as \(g\).
Let's first understand the formula for acceleration due to gravity. The acceleration due to gravity \(g\) on the surface of a celestial body is given by the formula:
\[ g = \frac{GM}{R^2} \]Where:
- \(G\) is the universal gravitational constant
- \(M\) is the mass of the celestial body
- \(R\) is the radius of the celestial body
Understanding Earth's Surface Gravity
For Earth, the acceleration due to gravity on its surface is given as \(g\). Using the formula, we can write this as:
\[ g_{\text{Earth}} = g = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \]Here, \(M_{\text{Earth}}\) is the mass of Earth and \(R_{\text{Earth}}\) is the radius of Earth.
Applying the Formula to the Planet
Now, let's consider the planet described in the question. We are given:
- Mass of the planet, \(M_{\text{Planet}}\), is the same as the mass of Earth, so \(M_{\text{Planet}} = M_{\text{Earth}}\).
- Radius of the planet, \(R_{\text{Planet}}\), is twice the radius of Earth, so \(R_{\text{Planet}} = 2R_{\text{Earth}}\).
Let \(g_{\text{Planet}}\) be the acceleration due to gravity on the surface of this planet. Using the same formula:
\[ g_{\text{Planet}} = \frac{GM_{\text{Planet}}}{R_{\text{Planet}}^2} \]Substitute the given values for \(M_{\text{Planet}}\) and \(R_{\text{Planet}}\) in terms of Earth's mass and radius:
\[ g_{\text{Planet}} = \frac{G(M_{\text{Earth}})}{(2R_{\text{Earth}})^2} \] \[ g_{\text{Planet}} = \frac{GM_{\text{Earth}}}{4R_{\text{Earth}}^2} \]Comparing Planet's Gravity to Earth's Gravity
We can rewrite the expression for \(g_{\text{Planet}}\) by separating the factor of \(1/4\):
\[ g_{\text{Planet}} = \frac{1}{4} \left( \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} \right) \]From our earlier equation for Earth's gravity, we know that \(g_{\text{Earth}} = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} = g\).
Substitute \(g\) into the equation for \(g_{\text{Planet}}\):
\[ g_{\text{Planet}} = \frac{1}{4} g \]So, the acceleration due to gravity on the surface of the planet is \(g/4\).
Summary of Calculation Steps
Here's a quick summary of the steps taken:
- Write down the formula for acceleration due to gravity: \(g = \frac{GM}{R^2}\).
- Write down the expression for Earth's gravity: \(g = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2}\).
- Write down the expression for the planet's gravity using its mass and radius: \(g_{\text{Planet}} = \frac{GM_{\text{Planet}}}{R_{\text{Planet}}^2}\).
- Substitute the given conditions: \(M_{\text{Planet}} = M_{\text{Earth}}\) and \(R_{\text{Planet}} = 2R_{\text{Earth}}\).
- Simplify the expression for \(g_{\text{Planet}}\): \(g_{\text{Planet}} = \frac{G M_{\text{Earth}}}{(2R_{\text{Earth}})^2} = \frac{G M_{\text{Earth}}}{4R_{\text{Earth}}^2}\).
- Relate the planet's gravity expression back to Earth's gravity expression: \(g_{\text{Planet}} = \frac{1}{4} \left( \frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2} \right) = \frac{1}{4} g\).
Conclusion on Planet's Surface Gravity
The acceleration due to gravity on the surface of the planet is one-fourth of the acceleration due to gravity on the surface of Earth. Therefore, if Earth's surface gravity is \(g\), the planet's surface gravity is \(g/4\).
| Property | Earth | Planet |
|---|---|---|
| Mass | \(M_{\text{Earth}}\) | \(M_{\text{Planet}} = M_{\text{Earth}}\) |
| Radius | \(R_{\text{Earth}}\) | \(R_{\text{Planet}} = 2R_{\text{Earth}}\) |
| Surface Gravity | \(g_{\text{Earth}} = g = \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2}\) | \(g_{\text{Planet}} = \frac{GM_{\text{Planet}}}{R_{\text{Planet}}^2} = \frac{G(M_{\text{Earth}})}{(2R_{\text{Earth}})^2} = \frac{1}{4} \frac{GM_{\text{Earth}}}{R_{\text{Earth}}^2} = \frac{g}{4}\) |
Revision Table: Key Concepts of Gravity
| Concept | Description | Formula |
|---|---|---|
| Gravitational Force | The attractive force between any two objects with mass. | \(F = \frac{Gm_1 m_2}{r^2}\) |
| Acceleration Due to Gravity (\(g\)) | The acceleration experienced by an object due to the gravitational force of a large body (like a planet or star). It is independent of the object's mass. | \(g = \frac{GM}{R^2}\) (on surface) \(g' = \frac{GM}{r^2}\) (at distance r from center, r > R) |
| Factors Affecting Surface Gravity | Mass (\(M\)) of the body (gravity is directly proportional to mass) Radius (\(R\)) of the body (gravity is inversely proportional to the square of the radius) |
\(g \propto M\) \(g \propto \frac{1}{R^2}\) |
Additional Information: How Mass and Radius Influence Gravity
The formula \(g = \frac{GM}{R^2}\) clearly shows how the acceleration due to gravity depends on the mass and radius of a celestial body.
- Effect of Mass: If the mass \(M\) increases while the radius \(R\) remains constant, the acceleration due to gravity \(g\) increases proportionally. This means more massive planets have stronger surface gravity (assuming similar density or radius).
- Effect of Radius: If the radius \(R\) increases while the mass \(M\) remains constant, the acceleration due to gravity \(g\) decreases with the square of the radius (\(1/R^2\)). This is because the surface is farther from the center of mass. In our problem, the radius doubled (\(2R_e\)), so the gravity became \(1/(2R_e)^2 = 1/(4R_e^2)\) times the original gravity factor related to radius.
In the given problem, the mass stayed the same, but the radius doubled. The effect of doubling the radius (from \(R_e\) to \(2R_e\)) while keeping the mass constant reduces the gravity by a factor of \(1/(2)^2 = 1/4\). So, the new gravity is \(g/4\).
Important Questions from Gravity
Which of the following statement is correct?
I. Gravitation is a weak force unless large masses are involved.
II. The weight is equal to the product of mass and acceleration due to gravity.
The mass of the Earth is ________.
What is the reason that the value of g (acceleration due to gravity) becomes greater at the poles than at the equator?
A man's mass is 72 kg on the earth. His mass on the moon will be:
The weight of 6 kg mass of a body on the surface of moon is about (for earth, \(g = 10 m/s^2 \) )
