Density of a planet is two times the density of earth. Radius of this planet is half. Match the following (as compared to earth)
`{:(,"Column-I",,"Column-II"),("(A)","Accleration due to gravity on this planet's surface","(p)","Half"),("(B)","Gravitational potential on the surface","(q)","Same"),("(C)","Gravitational potential at centre","(r)","Two times"),("(D)","Gravitational field strength at centre","(s)","Four times"):}`

Explanation

To solve the problem, we need to analyze the relationships between the parameters given for the planet and Earth.

### Given:
- Density of the planet, \( \rho_p = 2 \rho_e \) (where \( \rho_e \) is the density of Earth)
- Radius of the planet, \( r_p = \frac{1}{2} r_e \) (where \( r_e \) is the radius of Earth)

### Step 1: Calculate the acceleration due to gravity on the planet's surface (\( g_p \))

The formula for acceleration due to gravity is given by:

\[ g = \frac{G M}{r^2} \]

Where \( M \) is the mass of the planet. The mass can be expressed in terms of density and volume:

\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi r^3 \]

For the planet:

\[ M_p = \rho_p \cdot \frac{4}{3} \pi r_p^3 = (2 \rho_e) \cdot \frac{4}{3} \pi \left(\frac{1}{2} r_e\right)^3 \]

Calculating \( M_p \):

\[ M_p = (2 \rho_e) \cdot \frac{4}{3} \pi \cdot \frac{1}{8} r_e^3 = \frac{1}{3} \rho_e \cdot 4 \pi r_e^3 = \frac{1}{3} M_e \]

Now substituting \( M_p \) into the formula for \( g_p \):

\[ g_p = \frac{G M_p}{r_p^2} = \frac{G \cdot \frac{1}{3} M_e}{\left(\frac{1}{2} r_e\right)^2} \]

Calculating \( g_p \):

\[ g_p = \frac{G \cdot \frac{1}{3} M_e}{\frac{1}{4} r_e^2} = \frac{4}{3} \cdot \frac{G M_e}{r_e^2} = \frac{4}{3} g_e \]

### Step 2: Gravitational potential on the surface of the planet (\( V_p \))

The gravitational potential \( V \) is given by:

\[ V = -\frac{G M}{r} \]

For the planet:

\[ V_p = -\frac{G M_p}{r_p} = -\frac{G \cdot \frac{1}{3} M_e}{\frac{1}{2} r_e} \]

Calculating \( V_p \):

\[ V_p = -\frac{2}{3} \cdot \frac{G M_e}{r_e} = -\frac{2}{3} V_e \]

### Step 3: Gravitational potential at the center of the planet

The gravitational potential at the center of a uniform sphere is given by:

\[ V_c = -\frac{3}{2} \frac{G M}{r} \]

For the planet:

\[ V_{c,p} = -\frac{3}{2} \cdot \frac{G M_p}{r_p} = -\frac{3}{2} \cdot \frac{G \cdot \frac{1}{3} M_e}{\frac{1}{2} r_e} \]

Calculating \( V_{c,p} \):

\[ V_{c,p} = -\frac{3}{2} \cdot \frac{2}{3} \cdot \frac{G M_e}{r_e} = -\frac{G M_e}{r_e} = V_e \]

### Step 4: Gravitational field strength at the center of the planet

The gravitational field strength \( g_c \) at the center of a uniform sphere is zero.

### Summary of Results:
- \( g_p = \frac{4}{3} g_e \) (not listed in options)
- \( V_p = -\frac{2}{3} V_e \) (not listed in options)
- \( V_{c,p} = V_e \)
- \( g_c = 0 \)

### Matching:
- (A) \( g_p \) → (p) Half (Incorrect)
- (B) \( V_p \) → (q) Same (Incorrect)
- (C) \( V_{c,p} \) → (r) Two times (Incorrect)
- (D) \( g_c \) → (s) Four times (Incorrect)

### Final Matching:
- (A) → (p) Incorrect
- (B) → (q) Incorrect
- (C) → (r) Incorrect
- (D) → (s) Incorrect