Question

Gravitational pressure is the pressure that develops in solid objects due to the gravitational forces between different regions inside the object. If the earth is a uniform sphere with density $\rho = 5.5 \times 10^3 kg/m^3$ and radius $R = 6400km$, find the approximate gravitational pressure at its centre.

• A. $1.9 \times 10^8$ atm
• B. $1.7 \times 10^6$ atm
• C. $2.1 \times 10^3$ atm
• D. $0.6 \times 10^5$ atm

Answer 1

Answer: (B)

1.7 x 10^6 atm

Answer 2

To find the gravitational pressure at the Earth's center, we can model the Earth as a uniform sphere. The pressure at a distance $r$ from the center of a uniform sphere of radius $R$ and density $\rho$ is given by the formula derived from hydrostatic equilibrium:

$P(r) = \frac{2}{3}\pi G \rho^2 (R^2 - r^2)$

At the center of the Earth, $r=0$. So, the pressure at the center $P(0)$ is:

$P(0) = \frac{2}{3}\pi G \rho^2 R^2$

Given values:

• Density $\rho = 5.5 \times 10^3 \text{ kg/m}^3$
• Radius $R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}$
• Gravitational constant $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$

Substitute the values into the formula:

$P(0) = \frac{2}{3} \times \pi \times (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (5.5 \times 10^3 \text{ kg/m}^3)^2 \times (6.4 \times 10^6 \text{ m})^2 \approx 1.7 \times 10^6 \text{ atm}$