SSC 10th Science Maharashtra Board: Gravitation (Numerical Problem 5)
Question 5 (a)
An object takes $5\text{ s}$ to reach the ground from a height of $5\text{ m}$ on a planet. What is the value of $g$ on the planet?
Solution:
Given:
Initial velocity ($u$) = $0\text{ m/s}$ (since the object is falling)
Time ($t$) = $5\text{ s}$
Displacement ($s$) = $5\text{ m}$To find:
Acceleration due to gravity ($g$) = ?Formula:
Using Newton's second equation of motion:
$$s = ut + \frac{1}{2}gt^2$$Calculation:
Substitute the given values into the formula:$$5 = (0 \times 5) + \frac{1}{2} \times g \times (5)^2$$
$$5 = 0 + \frac{1}{2} \times g \times 25$$
$$5 = \frac{25}{2}g$$
$$g = \frac{5 \times 2}{25}$$
$$g = \frac{10}{25} = 0.4\text{ m/s}^2$$
The value of $g$ on the planet is $0.4\text{ m/s}^2$.
Question 5 (b)
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 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 $g_A = 2g_B$ and $R_B = 2R_A$:
$$\frac{2g_B}{g_B} = \frac{M_A}{M_B} \times \left(\frac{2R_A}{R_A}\right)^2$$
$$2 = \frac{M_A}{M_B} \times (2)^2$$
$$2 = \frac{M_A}{M_B} \times 4$$
$$M_B = \frac{4M_A}{2}$$
$$M_B = 2M_A$$
The mass of planet B must be $2M_A$.
