Tag: circular motion and gravitation

Questions Related to circular motion and gravitation

A body is acted upon by a constant force directed towards a fixed point. The magnitude of the force varies inversely as the square of the distance from the fixed point then path can be described by an equation similar to:

  1. $y=mx+c$

  2. ${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$

  3. $y=c{ x }^{ 2 }$

  4. none of these


Correct Option: D
Explanation:

The force field is similar to gravitational field.
So, the path is parabolic or elliptical or hyperbolic.
The exact trajectory will depend on the masses and  energy of the particle.
Option(A) is equation of a straight line, it is ruled out.
Option(B) is equation for a circle and is also eqn of an ellipse with length of major axis equal to minor axis.
Option(C) is eqn of a parabola.
Hence, (D) is the best option.

Potantial (V) at a point in space is given by $v = x^2 + y^2 + z^2$. Gravitational field at a point (x, y, z) is 

  1. $-2 x \hat{i} - 2 y \hat{j} - 2 z \hat{k}$

  2. $2 x \hat{i} + 2 y \hat{j} + 2 z \hat{k}$

  3. $x \hat{i} + y \hat{j} - z \hat{k}$

  4. $-x \hat{i} - y \hat{j} - z \hat{k}$


Correct Option: A
Explanation:

$v=x^2+y^2+z^2$

$E = \dfrac{{ - dv}}{{dx}}$

$E =  - \left[ {2\hat x + 2y\hat j + 2z\hat k} \right]$

$E =  - 2\hat x - 2y\hat j - 2z\hat k$
So, option $A$ is correct.

A large object is placed at exactly $65$% of the distance to the moon from the earth. Find out correct statement about the object ?

  1. Fall to the sun

  2. Fall to the moon

  3. Fall to the earth

  4. Remain in the same place

  5. Drift out of the solar system


Correct Option: C
Explanation:

The mass of moon is $1.2$% of that of earth. Hence $M _{moon}=0.012M _{earth}$

The acceleration due to gravity of the earth=$\dfrac{GM _{earth}}{(0.65R)^2}$
The acceleration due to gravity of the moon=$\dfrac{GM _{moon}}{(0.35R)^2}$
Hence $g _{earth}>g _{moon}$, and thus object falls towards the earth.

What is the time period satellite near the earth surface (neglect the height of orbit of satellite from the surface of ground)?

  1. $30.53 \ minutes$

  2. $50.38 \ minutes$

  3. $52.68 \ minutes$

  4. $84.75 \ minutes$


Correct Option: D
Explanation:

$T =$ Time period

$R=$ radius of earth $= 6400000m$
Gravitational constant $= G=$ $ 6.67\times {10}^{-11} Nm^2/kg^2$
$M=$ mass of earth $=$ $6\times{10}^{24}kg$

$T=$ $2\pi\sqrt{\dfrac {R^3}{GM}}$
Substitute and calculate
Whatever answer you get, divide it by $60$ so that it gets converted into minutes.

Two block of mass $10 kg$ and $20kg$ is separated by a distance $100 km$. What is the gravitational field (F) if $G=6.674 08 \times  10^{-11} m^3 kg^{-1} s^{-2}$?

  1. $13.34 \times 10^{-16} N$

  2. $1.334 \times 10^{-11} N$

  3. $ 1334 \times 10^{-16} N$

  4. $1.334 \times 10^{-18} N$


Correct Option: D
Explanation:
Given : $m _1 = 10kg$  $m _2 = 20 kg$  $r =10^5 m$
Gravitational force between the blocks $F = \dfrac{G m _1 m _2}{r^2}$
$\therefore$ $F = \dfrac{6.67408 \times 10^{-11} \times 10\times 20}{(10^5)^2}$
$\implies$ $1.334 \times 10^{-18} N$

If the radius of the earth were to shrink and its mass were to remain the same, the acceleration due to gravity on the surface of the earth with?

  1. Increase

  2. Decrease

  3. Remains same

  4. Zero


Correct Option: A
Explanation:

Gravitational force between two masses = $\quad K\dfrac { m1m2 }{ { r }^{ 2 } } \quad \quad $

Where

K is universal gravitational constant $m _1$ and $m _2$ are the masses $r$ is the distance between the masses.

Now , the acceleration due to $ m1 $ being earth will be = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad \quad $

If the radius decreases and the mass remains same the acceleration will increase.

What is the gravitational field strength at the surface of Jupiter (mass $1.9\times 10^{27} kg$, radius $7.1\times 10^7 m$?

  1. $25 N kg^{-1}$

  2. $25 N kg^{-2}$

  3. $35 N kg^{-1}$

  4. $55 N kg^{-2}$


Correct Option: A
Explanation:

Gravitational field = $ K\dfrac { m1 }{ { r }^{ 2 } } \quad $

where $K$ is universal gravitational constant $m _1$ is the mass $r$ is the distance. 

G$ 6.67\times { 10 }^{ -11 }\dfrac { 1.9({ 10 }^{ 27 }) }{ { (7.1\times { 10 }^{ 7 }) }^{ 2 } } \quad \quad $ 

   =  $\quad 25N{ kg }^{ -1 }\quad $

Both earth and moon are subject to the gravitational force to the sun. As observed from the sun, the orbit of the moon 

  1. will be elliptical.

  2. will not be strictly elliptical because the total gravitational force on it is not central.

  3. is not elliptical but will necessarily be closed curve.

  4. deviates considerably from being elliptical due to influence of plants other than earth.


Correct Option: B
Explanation:

If the moon were to experience only the gravitation of the earth only then the path would have been elliptical but in the present case moon experience the gravitational force of sun as well which distorts the path.

The fours basic forces in nature are.
I. Gravitational force
II. Electromagnetic force
III. Strong nuclear force
IV. Weak nuclear force
The relative magnitudes of these forces are in the order of.

  1. III $>$ II $>$ I $>$ IV

  2. III $>$ II $>$ IV $>$ I

  3. I $>$ II $>$ III $>$ IV

  4. III $>$ I $>$ II $>$ IV


Correct Option: B
Explanation:
Actually the order of the four basic forces in nature in order of their relative magnitudes are strong nuclear force $>$ Electromagnetic force $>$ Weak nuclear force $>$ Gravitational force.

Read the following statements.
I. A magnetic field is unable to penetrate into a superconductor.
II. An electric field can be cut off by a screen of conducting material
III. Gravitational field can be freely transmitted through all bodies
Which statement is false?

  1. III

  2. II

  3. I

  4. None of these


Correct Option: D
Explanation:

A. stronout feels weightless inside an artificial satellite because nothing is stopping their fall due to force of gravity.