Gravitation MCQs for NEET — Physics Questions with Answers

(b)     Given that, $T_1=1$ day and $T_2=8$ days

          $\therefore \frac{T_2}{T_1}={\left(\frac{r_2}{r_1}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow \frac{r_2}{r_1}={\left(\frac{T_2}{T_1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}={\left(\frac{8}{1}\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=4\Rightarrow r_2=4r_1=4R$

(c)    Time period of revolution of moon around the earth = 1 lunar month.

        $\frac{T_s}{T_m}={\left(\frac{r_s}{r_m}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}={\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\Rightarrow T_s=2^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ lunar month.

(d) Mechanical energy = Kinetic energy + potential energy

$$\begin{gathered} \Rightarrow \frac{1}{2}m{v}^2+\left(\frac{-GMm}{r}\right) \\ =\frac{1}{2}m\left(\frac{GM}{r}\right)-\frac{GMm}{r} \left[as v=\sqrt{\frac{GM}{r}}\right] \end{gathered}$$

Hence, mechanical energy = $\frac{-GMm}{2r}$

(d) Thinking process $g_h$ = Acceleration due to gravity at height above earth's surface 

      =$g\left(\frac{R}{R+h}\right)$

     =$g\left(1-\frac{2h}{R}\right)$

$g_d$=Acceleration at depth d below earth's surface 

       =$g\left(1-\frac{d}{R}\right)$

Given, when h=1km, $g_d$=$g_h$

or      $g\left(1-\frac{d}{R}\right)$=g$\left(1-\frac{2h}{R}\right)$

$\Rightarrow$  d=2h

0r   d=2km

(b) In the space, there is no external gravity. Due to masses of the astronauts , there will be small gravitational attractive force between them. Thus, these astronauts will move towards each other.

$$\begin{gathered} \left(b\right) Total energy of a satellite at height h=KE+PE \\ =\frac{GMm}{2\left(R+h\right)}-\frac{GMm}{(R+h)} \end{gathered}$$

$=\frac{-GMm}{2\left(R+h\right)}=\frac{-GMm{R}^2}{2R^2\left(R+h\right)}$

  =$\frac{-m{g}_o{R}^2}{2\left(R+h\right)} \left(\because g_o=\frac{GM}{R^2}\right)$

(d) Gravitational potential at some height h from the surface of the earth is given by

                   $V=-\frac{GM}{R+h}$              ...(i)

And accleration due to gravity at some height h from the earth surface can be given as

              $g\text{'}=\frac{GM}{{\left(R+h\right)}^2}$              ...(ii)

From eq.(i) and (ii), we get 

  $\frac{\left|V\right|}{g\text{'}}=\frac{GM}{\left(R+h\right)}\times \frac{{\displaystyle {\left(R+h\right)}^2}}{GM}$  

  $\Rightarrow \frac{\left|V\right|}{g\text{'}}=R+h$                          ...(iii)

$\because V=5.4\times {10}^{7}J k{g}^{-2}and g\text{'}=6.0 m{s}^{-2}$

Radius of earth, R=6400 km.

Substitute these values in eq. (iii), we get 

      $$\begin{gathered} \frac{5.4\times {10}^7}{6.0}=R+h\Rightarrow 9\times {10}^6=R+h \\ \Rightarrow h=\left(9-6.4\right)\times {10}^6=2.6\times {10}^{6}m \\ \Rightarrow h=2600 km \end{gathered}$$

(a) Since, the escape velocity of earth can be given as

  $\left(v_e\right)$=$\sqrt{2gR}=R\sqrt{\frac{8}{3}\mathrm{\pi G}\rho }$   $\left[\rho =density of earth\right]$

  $\Rightarrow v_e=R\sqrt{\frac{8}{3}\mathrm{\pi G\rho }}$         ...(i)

As it is given that the radius and mean density of planet are twice as that of earth. So, escape velocity at planet will be

 $v_p=2R\sqrt{\frac{8}{3}\mathrm{\pi G}2\rho }$          ...(ii) 

Divide, eq, (i) by eq.(ii), we get

      $$\begin{gathered} \frac{v_e}{v_p}=\frac{R\sqrt{\frac{8}{3}\mathrm{\pi G\rho }}}{2R\sqrt{\frac{8}{3}\mathrm{\pi G}\left(2\rho \right)}} \\ \Rightarrow \frac{v_e}{v_p}=\frac{1}{2\sqrt{2}} \end{gathered}$$

The gravitational force of attraction between the planet and sun provide the centripetal force
i.e.  $\frac{GMm}{r^2}$=mv²/r =>v=$\sqrt{\frac{GM}{r}}$

The time period of planet will be

T=2πr/v =>T²=$\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^2}{\mathrm{Gm}/\mathrm{r}}$=$\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}$...(i)

Also from Kepler's third law

T²=Kr³ ...(ii)

From Eqs. (i) and (ii), we get 
$\frac{4{\mathrm{\pi }}^2{\mathrm{r}}^3}{\mathrm{GM}}$=Kr³

=>GMK=4π²

The collision distance between two spherical bodies of masses M and 5M with initial separation 12R is 7.5R for the smaller body. This can be derived using Newton's law of gravitation and principles of conservation of energy and momentum.