Explanation
A body has weight due to the fall under the action of gravity. The unit of gravity is gal in C.G.S (named after Galileo). $$1 Gal =\dfrac{1cm}{sec^{2}}$$.
$$+300 mGal=0.3\times 10^{-2} ms^{-2}=0.003 m s^{-2}$$
$$\therefore g=9.78+0.003=9.783 m s^{-2}$$
Hint : Use the formula $$g'=\dfrac{GM}{(R+h)^{2}}$$
$$\textbf{Step1: Expression of acceleration due to gravity on earth's surface}$$
Force on the body placed on Earth's surface is$$F = \dfrac{GMm}{R^2}$$But, $$F = mg$$ hence,$$g = \dfrac{GMm}{R^2}$$where, variables have their usual meanings.$$\textbf{Step2: Calculation of altitude}$$Now, force on the body at geo-potential height say $$h$$ (altitude) where the acceleration due to gravity is $$25\%$$ of that at the earth's surface i.e.$$\dfrac{25 g}{100} = \dfrac{g}{4}$$
use the formula $$g'=\dfrac{GM}{(R+h)^{2}}$$Hence, we can write$$\dfrac{g}{4} = \dfrac{GM}{(R+h)^2}$$$$\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2}$$$$(R+h)^2 = 4R^2$$Taking roots for both sides we get$$R+h = 2R$$$$h = R$$
Hint: Use the Keplers law
Explanation:
According to Kepler's rules of planetary motion, when a satellite/planet moves around the Earth/Sun in a certain orbit, aerial velocity remains constant.
Correct option is (C) areal velocity
Velocity of the planet is $$v=\sqrt{\dfrac{GM}{a}}$$
Kinetic Energy is $$K=\dfrac{1}{2}m{v}^{2}=\dfrac{GMm}{2a}$$
Potential Energy is $$U=-{\dfrac{GMm}{a}}$$
$$Total\ energy= K+U= -{\dfrac{GMm}{2a}}$$
$${v}= \sqrt{\dfrac{GM}{a}}$$
So if$$,\quad v=1+0.414 v=1.414v= \sqrt{2} v= \sqrt{\dfrac{2GM}{a}}$$
Escape Velocity is $$v_e=\sqrt{\dfrac{2GM}{a}}$$
So the moon will escape.
A) At $$\infty$$ both $$V$$ and $$E$$ are zero.
B) Let, $$ V_ \infty = \dfrac{GM}{R} \rightarrow V_R = 0 $$ and $$ E_R = \dfrac{GM}{R^2}$$
C) Inside a spherical shell $$V =\dfrac{-GM}{R}$$ and $$E =0$$
D) Consider a point at a distance of $$r$$ from mass $$m$$. Then, $$V =\dfrac{-GM}{r}$$ and $$E =\dfrac{GM}{r^2}$$.
For, $$h<<R$$ force is constant and is equal to mg
Therefore, work done is mgh in moving by distance $$h$$
For, $$h=R$$
gravitational potential at x is $$-\dfrac { GM }{ x } $$
Work done in moving from $$x=R$$ to $$x=2R$$ is $$-\dfrac { GMm }{ 2R } +\dfrac { GMm }{ R } =\dfrac { GMm }{ 2R } $$
which is equal to $$\dfrac { 1 }{ 2 } mgR$$
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