Physics

Gravitation and Center of Mass

368 Questions

Gravitation and center of mass questions explore gravitational fields, planetary density, and the mechanics of celestial bodies. Test items include calculating gravitational strength on different planets and understanding the Roche Limit. This topic is essential for the physics syllabus of major competitive exams.

Gravitational fieldCenter of massPlanetary densityHill SphereSpace-time curvature

Gravitation and Center of Mass Questions

Multiple choice

What is a "g-force"?

  1. another word for "gravity"

  2. the gravitational force between earth and the moon

  3. a measure of acceleration

  4. A NASA crime-fighting unit
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A g-force is a measurement of acceleration expressed in multiples of the acceleration due to gravity on Earth.

Multiple choice

The gravity on _____________ is greater

  1. the moon

  2. space

  3. the earth

  4. in school
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Gravity is proportional to mass; Earth has significantly more mass than the moon, resulting in stronger gravity.

Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

An object is weighed in the following places using a spring balance. In which place will it weigh the heaviest?

  1. On the Moon

  2. At the equator

  3. At the pole

  4. In outer space
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Acceleration due to gravity $g$ is the largest at the pole of the earth.
Weight of the object $W = mg$
$\implies$ $W\propto g$
Thus the object will weigh the heaviest at the pole of the earth
Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

Two equal masses m and m are hung from a spring balance whose scale pans differ in vertical height by h. If $\rho$ is the mean density of earth ,then what is the error in weighing?

  1. $\frac { 8 } { 3 } \pi \mathrm { G\rho m h }$
  2. $\frac { 4 } { 3 } \pi \mathrm { G\rho m h }$
  3. $\frac { 5 } { 3 } \pi \mathrm { G\rho m h }$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The difference in gravitational potential due to the height difference h leads to a weight difference. The formula for the error in weighing due to gravity gradient is (8/3) * pi * G * rho * m * h.

Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

A body is suspended on a spring balance in ship sailing along the equator with a speed $V$. If $\omega$ is the angular speed of the earth and ${W} _{0}$ is the scale reading when the ship is at rest, the scale reading when the ship is sailing will be very close to:

  1. ${W} _{0}$
  2. ${W} _{0}(1+\cfrac{2\omega V}{g})$
  3. ${W} _{0}(1\pm\cfrac{2\omega V}{g})$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Coriolis force affects the apparent weight of an object on a moving ship. The change in weight depends on the direction of travel relative to the Earth's rotation, leading to the +/- term.

Multiple choice physics measurements and units measuring mass measurement of mass measuring instruments

Which of the following principle helps in the measurement of mass of the planets?

  1. Einstein's theory of relativity

  2. Newton's Law of gravitation

  3. Newton's Law of cooling

  4. Parallax Method
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

With the help of Newton's law of gravitation we can measure mass of planets.

Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

$F=\dfrac { G{ m } _{ 1 }{ m } _{ 2 } }{ { r }^{ 2 } } $   where $m _1$ amd $m _2$ are mass of planets, and $r$ is the distance between them  and $G$ is gravitation constant.
From this we can measure mass of planets.

Multiple choice potential energy of a system of charges potential energy of various configurations electrostatic potential and capacitance electrostatics physics

Two small spheres have mass ${m} _{1}$ and ${m} _{2}$ and hanging from massless insulating threads of lengths ${l} _{1}$ and ${l} _{2}$. Two spheres carry charges ${q} _{1}$ and ${q} _{2}$ respectively. The spheres hang such that they are on the same horizontal level and the threads are inclined to the vertical at angle ${\theta} _{1}$ and ${\theta} _{2}$ respectively. If $F _1 = F _2$, then:

  1. ${ \theta } _{ 1 }={ \theta } _{ 2 }$
  2. ${ M } _{ 1 }={ M } _{ 2 }$
  3. $\cfrac { l _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { l _{ 2 } }{ \tan { { \theta } _{ 2 } } } \quad $
  4. $\cfrac { q _{ 1 } }{ \tan { { \theta } _{ 1 } } } =\cfrac { q _{ 2 } }{ \tan { { \theta } _{ 2 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For sphere $1$

In equilibrium, from figure.
$\begin{array}{l} { T _{ 1 } }\cos { \theta _{ 1 } } ={ M _{ 1 } }g; \ { T _{ 1 } }\sin { \theta _{ 1 } } ={ F _{ 1 } } \ \therefore \tan { \theta _{ 1 } } =\dfrac { { { F _{ 1 } } } }{ { { M _{ 1 } }g } } . \end{array}$
For sphere $2$
In equilibrium, from figure.
$\begin{array}{l} { T _{ 2 } }\cos { \theta _{ 2 } } ={ M _{ 2 } }g; \ { T _{ 2 } }\sin { \theta _{ 2 } } ={ F _{ 2 } } \ \therefore \tan { \theta _{ 2 } } =\dfrac { { { F _{ 2 } } } }{ { { M _{ 2 } }g } } . \end{array}$
Force of repulsion between two charges are same
$\therefore F _1=F _2$
$\theta _1=\theta _2$ only if $\dfrac{{{F _1}}}{{{M _1}g}} = \dfrac{{{F _2}}}{{{M _2}g}}.$
But $F _1=F _2$, then $M _1=M _2$.

Multiple choice physics units and measurement: error analysis rounding off digits rounding of digits standard form

Find the order of magnitude of the mass of the star, whose radius is $384 \times 10^6$ m and average density is $4 \times 10^3$ kg $m^{-3}$ :

  1. 30

  2. 29

  3. 24

  4. 21
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Radius of star $r=384\times10^6m$

So, Volume of star is $V=4\pi r^3/3=2.371\times10^{26}m^3$
Density of star is $\rho=4\times10^3kgm^{-3}$
Mass of star is $M=V\times\rho\approx9.5\times10^{29}kg=0.95\times10^{30}kg$
So the mass is of order of $10^{30}kg$

Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

The radius of the earth is $6.37 \times 10^6 m$ and its mass is $5.975 \times 10^{24} kg$. Find the earth's average density to appropriate significant figures.

  1. $5 \times 10^3 kg m^{-3}$
  2. $5.52 \times 10^3 kg m^{-3}$
  3. $2 \times 10^3 kg m^{-3}$
  4. $5.52 \times 10^3 kg m^{-4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density = Mass / Volume. Volume = (4/3) * pi * r^3. Density = 5.975e24 / ((4/3) * pi * (6.37e6)^3) = 5518 kg/m^3, which is 5.52e3 kg/m^3.

Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

A force of 10 N of gravitational force in CGS units is  represented as

  1. $10 Dynes$
  2. $10^2 Dynes$
  3. $10^5 Dynes$
  4. $10^6 Dynes$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A force of 10 N of gravittional force in CGS units is represented by 10 Dynes.

Dyne is a CGS unit of force which accelerates a mass of one gram at the rate of one centimeter per second.

Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

The force experienced by a unit mass at a point in the gravitational field is called its

  1. gravitational intensity

  2. electric intensity

  3. magnetic intensity

  4. gravitational constant
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force experienced by a $unit $ mass at a point in the gravitational field $=F _g=mg=1\times g=g $ $ Newton $


We know that $g$ is called as gravitational intensity 

Hence correct answer is option $A $ 

Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

The ratio of SI units to CGS of the gravitational intensity is

  1. $10^3:1$

  2. infinity

  3. zero
  4. $10^2:1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gravitational intensity is nothing but the acceleration due to gravity(g)

SI unit of g is $m/s^2=100cm/s^2$
But CGS unit of g is $cm/s^2$ 
So SI unit:CGS unit $=100:1$ 

Hence correct answer is option $D $ 

Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

Value of gravitational constant, $'G'$ is

  1. $6.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$
  2. $4.674 08 \times 10^{-11} m^3 kg^{-1} s^{-2}$
  3. $6.674 08 \times 10^{11} m^3 kg^{-1} s^{-2}$
  4. $6.674 08 \times 10^{-11} m^3 kg^{-2} s^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The value of G in SI units is $6.67408\times {10^{-11}}$

And the unit is $ Newton.metre^2/kg^2 $
Newton is $kg.m.s^{-2} $
Hence it becomes $ m^3.kg^{-1}.s^{-2} $

Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

Gravitational field is 

  1. directly proportional to square of the distance between two masses.

  2. inversely proportional to square of the distance between two masses.

  3. directly proportional to the distance between two masses.

  4. inversely proportional to the distance between two masses.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The gravitational field is the region in which the gravitational force can be experienced or its presence can be felt. The intensity of gravitational field is the force acting on a unit mass of a body. 

Gravitational force is given by $Force = \dfrac{GMm} {R^2}$, 
where $G$ is gravitational constant, $M$ is the mass of the body which is creating the gravitational force, $m$ is the mass of the body which is undergoing gravitational force and $R$ is the distance between them. You can see that this force is inversely proportional to the square of the distance between the two masses. Gravitational field is given by gravitational force / mass of the 2nd body i.e. $m$. 
$Field = \dfrac{GM}{R^2}$. Hence it is inversely proportional to square of the distance between two masses.