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Question

Imagine that there exists a planet that has half the radius of the earth and half its mass. Then, the acceleration due to gravity on the surface of that planet in relation to the acceleration due to gravity g on the surface of the earth is

This question was previously asked in
RRB ALP/Technician 2018 CBT 2 Electronics Mechanic Question Paper (23-Jan-2019) (Shift 1)
The correct answer is
$G^1 = 2g$

Gravity Calculation on a New Planet

The acceleration due to gravity (\(g\)) on the surface of a celestial body is determined by its mass (\(M\)) and radius (\(R\)) using the formula:

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

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

Comparing Planet Gravity to Earth

Let \(g\) represent the acceleration due to gravity on Earth, with mass \(M_E\) and radius \(R_E\). The formula is:

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

For the new planet, let its acceleration due to gravity be \(g^1\), its mass be \(M_p\), and its radius be \(R_p\).

We are given the following relationships:

  • Mass of the planet: \(M_p = \frac{M_E}{2}\)
  • Radius of the planet: \(R_p = \frac{R_E}{2}\)

Using the gravity formula for the new planet:

\(g^1 = \frac{GM_p}{R_p^2}\)

Substitute the given values for \(M_p\) and \(R_p\) into the equation:

\(g^1 = \frac{G \left( \frac{M_E}{2} \right)}{\left( \frac{R_E}{2} \right)^2}\)

Simplify the expression step-by-step:

\(g^1 = \frac{G \cdot M_E / 2}{R_E^2 / 4}\)

\(g^1 = \frac{G M_E}{2} \times \frac{4}{R_E^2}\)

\(g^1 = 2 \times \frac{GM_E}{R_E^2}\)

Recognize that \(\frac{GM_E}{R_E^2}\) is equal to Earth's gravity, \(g\). Substitute \(g\) back into the equation:

\(g^1 = 2g\)

Conclusion on Planet Gravity

The acceleration due to gravity on the surface of this new planet is twice the acceleration due to gravity on Earth.

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Similar Questions

  1. The relation between acceleration due to gravity (\(g\)) and universal gravitational constant (\(G\)) is given by:
  2. If three particles, each of mass \(M\) are placed at the three corners of an equilateral triangle of side \(a\), the force exerted by this system on another particle of mass \(M\) placed at the midpoint of a side is

Important Questions from Universal law of gravitation

  1. Which of the following is NOT correct ?

  2. Three point masses each of mass m are placed at the three corners of an equilateral triangle of side x. Find the resultant force acting on any one particle at the corner.

  3. Which of the following is NOT correct ?

  4. Which phenomenon, successfully explained by the universal law of gravitation, relies on the combined and differential gravitational attraction exerted by both the moon and the sun on the earth?
  5. The law of gravitation helped to prove that:
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