SSC 10th Science Maharashtra Board: Numerical Problem 2
Chapter 1 (Gravitation) - Question 5 (b)
Note: This is the 2nd numerical problem listed in the Gravitation chapter exercise.
The radius of planet A is half the radius of planet B. If the mass of A is $M_A$, what must be the mass of B so that the value of $g$ on B is half that of its value on A?
Solution:
Given:
Radius of planet A ($R_A$) = $\frac{1}{2} R_B \implies R_B = 2R_A$
Acceleration due to gravity on B ($g_B$) = $\frac{1}{2} g_A \implies g_A = 2g_B$
Mass of planet A = $M_A$To find:
Mass of planet B ($M_B$) = ?Formula:
The acceleration due to gravity on a planet's surface is given by:
$$g = \frac{GM}{R^2}$$Calculation:
For planet A: $$g_A = \frac{GM_A}{R_A^2}$$
For planet B: $$g_B = \frac{GM_B}{R_B^2}$$Taking the ratio of $g_A$ to $g_B$:
$$\frac{g_A}{g_B} = \frac{\frac{GM_A}{R_A^2}}{\frac{GM_B}{R_B^2}}$$
$$\frac{g_A}{g_B} = \frac{M_A}{M_B} \times \left(\frac{R_B}{R_A}\right)^2$$
Substitute the given conditions ($g_A = 2g_B$ and $R_B = 2R_A$) into the equation:
$$\frac{2g_B}{g_B} = \frac{M_A}{M_B} \times \left(\frac{2R_A}{R_A}\right)^2$$
$$2 = \frac{M_A}{M_B} \times (2)^2$$
$$2 = \frac{M_A}{M_B} \times 4$$
Rearranging to solve for $M_B$:
$$M_B = \frac{4M_A}{2}$$
$$M_B = 2M_A$$
Answer:
The mass of planet B must be $2M_A$ (twice the mass of planet A).
