The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is `9 : 4`. The mass of the planet is `(1)/(9) th` of that of the Earth, what is the radius of the planet ? (Tale the planets to have the same mass density).

Options

A. `(R)/(3)`
B. `(R)/(4)`
C. `(R)/(9)`
D. `(R)/(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between weight and gravitational acceleration
The weight of a body on the surface of a planet is given by the formula:
\[ W = mg \]
where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity on that planet.

### Step 2: Set up the ratio of weights
We are given that the ratio of the weights of a body on the Earth's surface to that on the surface of a planet is \( 9:4 \). Thus, we can write:
\[ \frac{W_E}{W_P} = \frac{9}{4} \]
This implies:
\[ \frac{mg_E}{mg_P} = \frac{9}{4} \]
Since the mass \( m \) is the same for both cases, we can simplify this to:
\[ \frac{g_E}{g_P} = \frac{9}{4} \]

### Step 3: Express gravitational acceleration in terms of mass and radius
The gravitational acceleration \( g \) at the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

For Earth and the planet, we can write:
\[ g_E = \frac{GM_E}{R_E^2} \]
\[ g_P = \frac{GM_P}{R_P^2} \]

### Step 4: Substitute the expressions for \( g_E \) and \( g_P \) into the ratio
Substituting these into our ratio gives:
\[ \frac{g_E}{g_P} = \frac{\frac{GM_E}{R_E^2}}{\frac{GM_P}{R_P^2}} = \frac{M_E R_P^2}{M_P R_E^2} \]

### Step 5: Use the given mass relationship
We know that the mass of the planet \( M_P \) is \( \frac{1}{9} M_E \). Substituting this into the equation gives:
\[ \frac{g_E}{g_P} = \frac{M_E R_P^2}{\frac{1}{9} M_E R_E^2} = \frac{9 R_P^2}{R_E^2} \]

### Step 6: Set up the equation using the weight ratio
From the weight ratio, we have:
\[ \frac{g_E}{g_P} = \frac{9}{4} \]
Thus, we can equate:
\[ \frac{9 R_P^2}{R_E^2} = \frac{9}{4} \]

### Step 7: Solve for \( R_P \)
Cancelling \( 9 \) from both sides, we get:
\[ \frac{R_P^2}{R_E^2} = \frac{1}{4} \]
Taking the square root gives:
\[ \frac{R_P}{R_E} = \frac{1}{2} \]
Thus, we find:
\[ R_P = \frac{R_E}{2} \]

### Final Answer
The radius of the planet is half the radius of the Earth.

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