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Question

The radii of two planets are respectively R 1and R 2and their densities are respectively ρ 1and ρ 2. The ratio of the acceleration due to gravity (g 1/g 2) at their surface is

The correct answer is \(\frac{R_1\rho_1}{R_2\rho_2}\)

Understanding Acceleration Due to Gravity

The question asks for the ratio of the acceleration due to gravity at the surface of two planets, given their radii and densities. The acceleration due to gravity on the surface of a planet depends on its mass and radius.

The general formula for the acceleration due to gravity ($g$) on the surface of a spherical body with mass $M$ and radius $R$ is given by:

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

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

Relating Mass and Density for a Planet

We are given the densities of the planets, not their masses directly. Assuming the planets are uniform spheres, the mass ($M$) of a planet can be related to its density (\(\rho\)) and volume ($V$). The volume of a sphere with radius \(R\) is \(V = \frac{4}{3}\pi R^3\). Therefore, the mass is:

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

Calculating Acceleration Due to Gravity Using Density and Radius

Now, we can substitute the expression for mass ($M$) in terms of density (\(\rho\)) and radius ($R$) into the formula for acceleration due to gravity:

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

Simplifying this expression, we get:

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

This formula shows that the acceleration due to gravity at the surface of a planet is directly proportional to its density and its radius.

Finding the Ratio of Acceleration Due to Gravity

We have two planets, Planet 1 and Planet 2, with radii \(R_1\), \(R_2\) and densities \(\rho_1\), \(\rho_2\), respectively.

Using the derived formula for acceleration due to gravity:

  • For Planet 1: \(g_1 = \frac{4}{3}\pi G \rho_1 R_1\)
  • For Planet 2: \(g_2 = \frac{4}{3}\pi G \rho_2 R_2\)

The ratio of the acceleration due to gravity (\(g_1/g_2\)) at their surfaces is:

\(\frac{g_1}{g_2} = \frac{\frac{4}{3}\pi G \rho_1 R_1}{\frac{4}{3}\pi G \rho_2 R_2}\)

We can cancel out the common terms \(\frac{4}{3}\pi G\) from the numerator and the denominator:

\(\frac{g_1}{g_2} = \frac{\rho_1 R_1}{\rho_2 R_2}\)

Rearranging the terms to match the options:

\(\frac{g_1}{g_2} = \frac{R_1\rho_1}{R_2\rho_2}\)

This is the ratio of the acceleration due to gravity for the two planets.

Conclusion

The ratio of the acceleration due to gravity at the surface of the two planets is found to be \(\frac{R_1\rho_1}{R_2\rho_2}\), based on their radii and densities. This result is derived by relating the acceleration due to gravity formula to the mass calculated from density and volume, highlighting the relationship between acceleration due to gravity, radius, and density.

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Important Questions from Acceleration due to gravity of the earth

  1. If a ball is thrown vertically upwards with the speed $u$, the distance covered during the second-to-last $t$ seconds of its ascent is
    (Assume $t$ is less than half of the total ascent time).
  2. A body freely falling from rest has acquired a velocity ‘v’ after it falls through a distance ‘h’. The distance it has to fall down further for its velocity to become double is:

  3. On earth, the value of G = 6.67 × 10 -11  Nm 2kg -2 . What is the value on moon, where acceleration due to gravity is nearly one - sixth than that of earth?

  4. What is the force required to produce an acceleration of 9.8 m/s 2on a body of weight 9.8N? Take g = 9.8 m/s 2.

  5. At what height above the surface of the earth does the weight of an object reduce by 1%. Given the radius of the earth is 6400.

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