Question

Imagine a new planet having the same density as that of earth, but it is 3 times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is \(g\) and that on the surface of the planet is \(g_{1}\), then \(g_{1}\) is equal to

• A. \(3g\)
• B. \(9g\)
• C. \(27g\)
• D. \(\frac{g}{9}\)

Hint:

For the same density, g is proportional to the radius.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
\(g=\dfrac{GM}{R^2}\) with \(M=\dfrac{4}{3}\pi R^3\rho\), so \(g=\dfrac{4}{3}\pi G\rho R\).

Step 2: Proportionality
With the same density, \(g\propto R\).

Step 3: Calculation
The planet's radius is 3 times that of the earth, so \(g_1=3g\).

Final Answer:
The acceleration due to gravity on the planet is 3g. \[ \boxed{\text{(A) }3g} \]