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). 1Gal=1cmsec2.
+300mGal=0.3×10−2ms−2=0.003ms−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 isF = \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^2Taking roots for both sides we getR+h = 2Rh = 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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