Question

The minimum and maximum distances of a satellite from the centre of the Earth are 2R and 4R respectively, where R is the radius of Earth and M is the Earth.
The radius of curvature at the point of minimum distance is
A $\quad \sqrt{\dfrac{8 R}{3}}$
B $\quad \sqrt{\dfrac{5 R}{3}}$
C $\quad \sqrt{\dfrac{4 R}{3}}$
D $\quad \sqrt{\dfrac{7 R}{3}}$

Answer 1

Hint We know that momentum is a physics term that refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum. One example is the use of air bags in automobiles. Air bags are used in automobiles because they are able to minimize the effect of the force on an object involved in a collision. Air bags accomplish this by extending the time required to stop the momentum of the driver and passenger. Momentum is mass in motion, and any moving object can have momentum. An object's change in momentum is equal to its impulse. Impulse is a quantity of force times the time interval. Impulse is not equal to momentum itself; rather, it's the increase or decrease of an object's momentum. Based on this concept we have to solve this question

Complete step by step answer
Using conservation of angular momentum which states that any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero. Angular momentum is just linear momentum affected by some tendency to turn just so an object remains the same distance away from a central point. Angular momentum is conserved because linear momentum is conserved. The distance something travels in a circular motion is simply the angle in radians times the radius.
$\mathrm{m}_{\text {satellite } \mathrm{v}_{1}} \mathrm{R}_{1}=\mathrm{m}_{\text {satellite } \mathrm{v}_{2}} \mathrm{R}_{2}$
$\mathrm{v}_{1} \mathrm{R}_{1}=\mathrm{v}_{2} \mathrm{R}_{2}$
$\mathrm{v}_{1} 2 \mathrm{R}=\mathrm{v}_{2} 4 \mathrm{R}$
$\mathrm{v}_{1}=2 \mathrm{v}_{2} \ldots$(i)
Using energy conservation equation:
$-\dfrac{\mathrm{GMm}}{2 \mathrm{R}}+\dfrac{1}{2} \mathrm{mv}_{1}^{2}=-\dfrac{\mathrm{GMm}}{4 \mathrm{R}}+\dfrac{1}{2} \mathrm{mv}_{2}^{2} \ldots$ (ii)
Substituting ${{v}_{2}}$ in equation (ii) from equation (i)
$-\dfrac{\mathrm{GMm}}{4 \mathrm{R}}=\dfrac{3}{2} \mathrm{mv}_{1}^{2}$
$\mathrm{v}_{1}=\sqrt{\dfrac{\mathrm{GM}}{6 \mathrm{R}}}$
$\mathrm{v}_{2}=\sqrt{\dfrac{2 \mathrm{GM}}{3 \mathrm{R}}}$
Radius of curvature $=\dfrac{\mathrm{v}^{2}}{\mathrm{a}_{\mathrm{n}}}=\dfrac{\left(\sqrt{\dfrac{\mathrm{GM}}{6 \mathrm{R}}}\right)^{2}}{\dfrac{\mathrm{GM}}{(2 \mathrm{R})^{2}}}$
So, by evaluating we get: $\quad \sqrt{\dfrac{8 R}{3}}$

So, the correct answer is option A.

Note We should know that linear momentum is defined as the product of a system's mass multiplied by its velocity. In symbols, linear momentum is expressed as $p=mv$. Momentum is directly proportional to the object's mass and also its velocity. Thus, the greater an object's mass or the greater its velocity, the greater its momentum. Note that the linear moment is a vector quantity and is conserved in any direction. In the center-of-mass system, the total momentum is always zero, before and after the interaction, in any direction. Angular momentum is inertia of rotation motion. Linear momentum is inertia of translation motion. The big difference is that the type of motion which is related to each momentum is different. It is important to consider the place where the force related to rotation applies, which appears as 'r' in the formula.