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 properties of the planet in comparison to Earth based on the given information: the density of the planet is two times that of Earth, and the radius of the planet is half that of Earth.

### Step-by-Step Solution:

1. **Acceleration due to Gravity (g') on the Surface of the Planet**:
The formula for acceleration due to gravity at the surface of a planet is given by:
\[ g' = \frac{G M}{R^2} \]
where \(M\) is the mass of the planet and \(R\) is its radius. The mass \(M\) can be expressed in terms of density (\(\rho\)) and volume (\(V\)):
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]
Substituting this into the equation for \(g'\):
\[ g' = \frac{G \rho \left(\frac{4}{3} \pi R^3\right)}{R^2} = \frac{4}{3} G \rho R \]
Given that the density of the planet is \(2\rho_E\) (where \(\rho_E\) is the density of Earth) and the radius is \(\frac{1}{2}R_E\):
\[ g' = \frac{4}{3} G (2\rho_E) \left(\frac{1}{2}R_E\right) = \frac{4}{3} G \rho_E R_E = g_E \]
Thus, the acceleration due to gravity on the planet's surface is the same as that of Earth.

2. **Gravitational Potential (V') on the Surface**:
The gravitational potential at the surface is given by:
\[ V' = -\frac{G M}{R} \]
Substituting for \(M\):
\[ V' = -\frac{G \left(2\rho_E \cdot \frac{4}{3} \pi \left(\frac{1}{2}R_E\right)^3\right)}{\frac{1}{2}R_E} \]
Simplifying:
\[ V' = -\frac{G \cdot 2\rho_E \cdot \frac{4}{3} \pi \cdot \frac{1}{8}R_E^3}{\frac{1}{2}R_E} = -\frac{G \cdot 2\rho_E \cdot \frac{4}{3} \pi \cdot \frac{1}{8}R_E^2}{\frac{1}{2}} = -\frac{G \cdot 2\rho_E \cdot \frac{4}{3} \pi \cdot \frac{1}{4}R_E^2}{1} = -\frac{G \cdot \rho_E \cdot \frac{4}{3} \pi R_E^2}{1} \]
This shows that the gravitational potential at the surface is the same as that of Earth.

3. **Gravitational Potential at the Center**:
The gravitational potential at the center of a solid sphere is given by:
\[ V_c = -\frac{3}{2} \frac{G M}{R} \]
For our planet:
\[ V_c' = -\frac{3}{2} \frac{G (2\rho_E \cdot \frac{4}{3} \pi \left(\frac{1}{2}R_E\right)^3)}{\frac{1}{2}R_E} = -\frac{3}{2} \cdot \frac{G \cdot 2\rho_E \cdot \frac{4}{3} \pi \cdot \frac{1}{8}R_E^3}{\frac{1}{2}R_E} \]
This simplifies to show that the gravitational potential at the center of the planet is two times that of Earth.

4. **Gravitational Field Strength (E) at the Center**:
The gravitational field strength inside a solid sphere is zero at the center. Therefore, the gravitational field strength at the center of the planet is also zero, just like on Earth.

### Final Matching:
- (A) Acceleration due to gravity on this planet's surface → (q) Same
- (B) Gravitational potential on the surface → (q) Same
- (C) Gravitational potential at center → (r) Two times
- (D) Gravitational field strength at center → (s) Zero