CBSE Class 11 Physics Chapter 8 Gravitation NCERT Solutions

Solution:

1. No, you cannot shield a body from the gravitational influence of nearby matter. Unlike electrical forces, which can be blocked by conductors, the gravitational force is independent of the medium and the nature of the objects involved. It is a fundamental force that permeates all space and cannot be screened by any material or structure.

2. Yes, if the space station is large enough, the astronaut might be able to detect gravity. While astronauts in small, freely falling spaceships experience weightlessness, in a very large space station, different parts of their body would be at slightly different distances from the Earth. This difference in distance would lead to a small but detectable difference in the gravitational force experienced, allowing for the detection of gravity.

3. The tidal effect of a celestial body depends not only on its gravitational force but also on the difference in this force across the Earth. This differential force is inversely proportional to the cube of the distance ($1/d^3$), whereas the direct gravitational force is inversely proportional to the square of the distance ($1/d^2$). Since the Moon is significantly closer to the Earth than the Sun, the inverse cube relationship causes the Moon's tidal effect to be greater, even though the Sun exerts a stronger overall gravitational pull on the Earth.

Solution:

a) Acceleration due to gravity decreases with increasing altitude.

b) Acceleration due to gravity decreases with increasing depth. (assuming the earth to be a sphere of uniform density).

c) Acceleration due to gravity is independent of the mass of the body.

d) The formula $-G\; Mm(1/r\_2-1/r\_1)$ is more accurate than the formula $mg(r\_2-r\_1)$ for the difference of potential energy between two points $r\_2$ and $r\_1$ distance away from the centre of the earth.

Explanation:

a) The acceleration due to gravity ($g$) at a height $h$ above the Earth's surface (radius $R\_e$) is given by:

$$g\_h\; =\; \backslash frac\{GM\}\{(R\_e\; +\; h)^2\}$$

Since $g\; =\; \backslash frac\{GM\}\{R\_e^2\}$, we can write $g\_h\; =\; g\; \backslash left(\backslash frac\{R\_e\}\{R\_e\; +\; h\}\backslash right)^2$. As $h$ increases, $R\_e\; +\; h$ increases, and thus $g\_h$ decreases.

b) For a point at depth $d$ inside the Earth (assuming uniform density), the acceleration due to gravity is given by:

$$g\_d\; =\; g\; \backslash left(1\; -\; \backslash frac\{d\}\{R\_e\}\backslash right)$$

As $d$ increases (meaning the depth increases), the term $\backslash frac\{d\}\{R\_e\}$ increases, and thus $g\_d$ decreases.

c) The acceleration due to gravity is given by $g\; =\; \backslash frac\{GM\}\{R^2\}$, where $M$ is the mass of the Earth and $R$ is its radius. This formula does not include the mass of the body ($m$) on which gravity acts. Therefore, $g$ is independent of the mass of the body.

d) The formula $mg(r\_2-r\_1)$ is an approximation that is valid only when the change in distance ($r\_2-r\_1$) is small compared to the distance from the Earth's center, and $g$ is assumed to be constant. The formula $-G\; Mm(1/r\_2-1/r\_1)$ represents the exact change in gravitational potential energy between two points at distances $r\_1$ and $r\_2$ from the center of the Earth, considering the variation of gravitational force with distance. Hence, it is more accurate.