NEET Practice Questions (MCQs) with Answers & Solutions

(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.

As we know that, force on satellite is only gravitational force which will always be towards the centre of earth Thus, the acceleration of S is always directed towards the centre of the earth