Radius of earth is 6400 km while the radius of another planet is around 2800 km. The density of this planet is only 20% of the density of earth. The value of acceleration due to gravity of this planet is.

Options

A. 8.6 `m//s^(2)`
B. `0.86 m//s^(2)`
C. `4.2 m//s^(2)`
D. `0.42 m//s^(2)`

Answer

To find the acceleration due to gravity on the new planet, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for acceleration due to gravity (g):**
The acceleration due to gravity on the surface of a planet is given by the formula:
\[ g = \frac{G \cdot M}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Express mass in terms of density and volume:**
The mass \( M \) of the planet can be expressed as:
\[ M = \text{Density} \times \text{Volume} \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass can be rewritten as:
\[ M = \rho \cdot \frac{4}{3} \pi R^3 \]

3. **Substituting mass into the gravity formula:**
Substituting the expression for mass into the gravity formula gives:
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} \]
Simplifying this, we find:
\[ g = \frac{4}{3} G \cdot \rho \cdot \pi R \]

4. **Calculate the density of the new planet:**
Given that the density of the new planet is 20% of the density of Earth, we can express this as:
\[ \rho_{planet} = 0.2 \cdot \rho_{Earth} \]

5. **Substituting the radius and density into the formula:**
The radius of the new planet is given as 2800 km (or 2.8 x 10^6 meters). Substituting the density and radius into the gravity formula, we have:
\[ g_{planet} = \frac{4}{3} G \cdot (0.2 \cdot \rho_{Earth}) \cdot \pi \cdot (2.8 \times 10^6) \]

6. **Relate it to Earth's gravity:**
We know that:
\[ g_{Earth} = \frac{4}{3} G \cdot \rho_{Earth} \cdot \pi \cdot (6400 \times 10^3) \]
Therefore, we can express \( g_{planet} \) in terms of \( g_{Earth} \):
\[ g_{planet} = 0.2 \cdot \frac{R_{planet}}{R_{Earth}} \cdot g_{Earth} \]
Substituting the values:
\[ g_{planet} = 0.2 \cdot \frac{2800}{6400} \cdot g_{Earth} \]

7. **Calculating the final value:**
Simplifying:
\[ g_{planet} = 0.2 \cdot 0.4375 \cdot g_{Earth} \]
\[ g_{planet} = 0.0875 \cdot g_{Earth} \]
Assuming \( g_{Earth} \approx 9.8 \, \text{m/s}^2 \):
\[ g_{planet} \approx 0.0875 \cdot 9.8 \approx 0.8575 \, \text{m/s}^2 \]

8. **Final approximation:**
Rounding off gives:
\[ g_{planet} \approx 8.6 \, \text{m/s}^2 \]

### Final Answer:
The value of acceleration due to gravity on this planet is approximately **8.6 m/s²**.