Q: The value of escape velocity from the surface of the Earth is approximately:

Using $ve = \sqrt{2gR}$ where $g = 9.8\text{ m/s}^2$ and $R = 6.4 \times 10^6\text{ m}$, we get $ve \approx 11.2\text{ km/s}$.

Question 1

Kepler's second law (law of areas) is a direct consequence of the law of conservation of:

Accepted Answer

Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. This is a direct consequence of the conservation of angular momentum.

Question 2

What is the dimensional formula of Universal Gravitational Constant ($G$)?

Accepted Answer

From $F = \frac{G m1 m2}{r^2}$, we get $G = \frac{F r^2}{m1 m2}$. Substituting dimensions: $G = \frac{M L T^{-2}L^2}{M^2} = M^{-1} L^3 T^{-2}$.

Question 3

The value of acceleration due to gravity ($g$) at the centre of the Earth is:

Accepted Answer

At a depth $d = R$ (centre of the Earth), $g' = g\left(1 - \frac{d}{R}\right) = g\left(1 - \frac{R}{R}\right) = 0$.

Question 4

The relation between escape velocity ($ve$) and orbital velocity ($vo$) for a satellite near the surface of Earth is:

Accepted Answer

Escape velocity is $ve = \sqrt{\frac{2GM}{R}}$ and orbital velocity is $vo = \sqrt{\frac{GM}{R}}$. Thus, $ve = \sqrt{2} vo$.

Question 5

The value of escape velocity from the surface of the Earth is approximately:

Accepted Answer

Using $ve = \sqrt{2gR}$ where $g = 9.8	ext{ m/s}^2$ and $R = 6.4 	imes 10^6	ext{ m}$, we get $ve \approx 11.2	ext{ km/s}$.

Gravitation — CBSE Class 11 Physics Important Questions & Revision Notes