Physics

Gravitation and Center of Mass

369 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 physics force and newton's laws of motion concept of inertia galileo's law and inertia mass and inertia

Suppose gravity of earth suddenly becomes zero, then in which direction
will the moon begin to move if no other celestial body affects it ?

  1. Radial

  2. Tangential

  3. cant say

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Moon will be launched in tangential direction with a speed at which it was orbiting around the earth.

In other words, it's speed and direction will be what it was at the time when gravitational force of earth ceased to exist.

Multiple choice physics sky vision types of galaxies galaxies constellations

Astronomers have observed a small massive object at the centre of our Milky Way galaxy. A ringof material orbits this massive object; the ring has a diameter of about 15 light years and an orbital  speed of 200$\mathrm { km } / \mathrm { s }$ . What is the mass of the massive object at the centre of the minky  Way? Given $G = 6.67 \times 10 ^ { - 11 } \mathrm { Nm } ^ { 2 } \mathrm { kg } ^ { 2 }$ and 1 light year $= 9.5 \times 10 ^ { 15 } \mathrm { m }$ 

  1. Approximately $8.1 \times 10 ^ { 37 } \mathrm { kg }$
  2. approximately $4.3 \times 10 ^ { 37 } \mathrm { kg }$
  3. Approximately $6 \times 10 ^ { 37 } \mathrm { kg }$
  4. approximately $3 \times 10 ^ { 37 } \mathrm { kg }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the formula for orbital velocity v = sqrt(GM/r), where r is the radius (half the diameter) and v is the speed, we can solve for M. Given diameter = 15 light years (r = 7.5 light years), v = 200,000 m/s, and the constants, the calculation yields approximately 4.3 x 10^37 kg.

Multiple choice physics dynamics - explaining motion system of unit summary of si units system of units

If 1 kg wt = 10 N, the value of gravitational intensity will be 

  1. $\displaystyle 10{ m }/{ { s }^{ 2 } }$
  2. $\displaystyle \frac { 1 }{ 10 } { m }/{ { s }^{ 2 } }$
  3. $\displaystyle 1{ m }/{ { s }^{ 2 } }$
  4. $\displaystyle \frac { 1 }{ 100 } { m }/{ { s }^{ 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If $1\,kg\,wt=10\,N=F$

Then gravitational intensity is given by,
$g=\dfrac Fm$
$=\dfrac{10 }{1}=10 m/s^2$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

The value of $g$ on the surface of earth is 9.8 $m / s ^ { 2 }$ and the radius of earth is $6400km$. The average density of earth in $k g / m ^ { 3 }$ will be

  1. $5.48 \times 10 ^ { 3 }$
  2. $2.64 \times 10 ^ { 3 }$
  3. $7.60 \times 10 ^ { 3 }$
  4. $1.46 \times 10 ^ { 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using g = G * M / R^2 and M = density * (4/3) * pi * R^3, we get g = G * density * (4/3) * pi * R. Solving for density = 3g / (4 * pi * G * R). Plugging in g=9.8, G=6.67e-11, R=6.4e6, we get approx 5.48e3 kg/m^3.

Multiple choice physics work and power commercial unit of energy power work and energy

A body projected vertically from the earth reaches a height equal to earth's radius before returning to the earth. The power exerted by the gravitational force is greatest :

  1. at the highest position of the body

  2. at the instant just before the body hits the earth

  3. it remains constant throughout

  4. at the instant just after the body is projected
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Power,$ $P$ $=$$\overrightarrow{F}$.$\overrightarrow{v}$ $=$ $Fv$ $cos$$\theta$
$Just$ $before$ $hitting$ $the$ $earth$ $θ$ $=$ $0°.$ $Hence,$ $the$ $power$ $exerted$ $by$ $the$ $gravitational$ $force$ $is$ $greatest$ $at$ $the$ $instant$ $just$ $before$ $the$ $body$ $hits$ $the$ $earth.$

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

What impulse need to be given to a body of mass $m$, released from the surface of earth along a straight tunnel passsing through centre of earth, at the centre of earth, to bring it to rest(Mass of earth $M$, radius of earth R) 

  1. $m \sqrt { \dfrac { G M } { R } }$
  2. $\sqrt { \dfrac { G M m } { R } }$
  3. $m \sqrt { \dfrac { G M } {2 R } }$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the center of the Earth, the gravitational force is zero. Since the body is already at the center and has reached it due to gravity, its velocity is at a maximum; however, the question asks for the impulse to bring it to rest. If the body is released from the surface, it will oscillate through the center. At the exact center, the net force is zero, but the body has kinetic energy. However, in the context of standard physics problems of this type, the force at the center is zero, and if we assume the body is meant to be at rest at the center, the impulse required is zero if it is already there, or the question implies a conceptual trick.

Multiple choice physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

$F _g$ and $F _e$ represent gravitational and electrostatic force respectively between electrons situated at a distance $0.1\,m. F _g/F _e$ is of the order

  1. $10^{-41}$
  2. $10^{-45}$
  3. $10^{40}$
  4. $10^{-42}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
${ F } _{ y }$ and ${ F } _{ e }$ represent gravitational and electrostatic force respectively between electrons situated at a distance $=0.1m$
let, us take the ratio
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { \dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \dfrac { { q }^{ 2 } }{ { d }^{ 2 } }  }{ G\dfrac { { m }^{ 2 } }{ { d }^{ 2 } }  } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =\dfrac { 1 }{ 4\pi { \epsilon  } _{ 0 } } \times \dfrac { 1 }{ G } \times \dfrac { { q }^{ 2 } }{ { m }^{ 2 } } $
$\dfrac { { F } _{ e } }{ { F } _{ g } } =9\times { 10 }^{ 9 }\times \dfrac { 1 }{ 6.67\times { 10 }^{ -11 } } \times \dfrac { { \left( 1.6\times { 10 }^{ -19 } \right)  }^{ 2 } }{ { \left( 9.1\times { 10 }^{ -31 } \right)  }^{ 2 } } $
$\Rightarrow \dfrac { { F } _{ e } }{ { F } _{ g } } =4.17\times { 10 }^{ 42 }$
or,  $\dfrac { { F } _{ e } }{ { F } _{ g } } $ is of the order of ${ 10 }^{ -42 }$.
Multiple choice physics magnetic fields and electromagnetism contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

The ratio of electric force between two electrons to the gravitational force between them is of the order

  1. 10$^{42}$
  2. 10$^{39}$
  3. 10$^{36}$

  4. 1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric force between two electrons is $F _e=\dfrac{ke^2}{r^2}$

and gravitational force between two electrons is $F _g=\dfrac{Gm _e^2}{r^2}$
Thus, $\dfrac{F _e}{F _g}=\dfrac{ke^2}{Gm _e^2}=\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{(6.67\times 10^{-11})(9.1\times 10^{-31})^2}=4.17\times 10^{42} \sim 10^{42}$

Multiple choice physics motion, force and speed contact and non-contact forces comparing force in magnetic, electric and gravitational fields identifying forces

Gravity is shown by

  1. earth only

  2. earth and moon

  3. earth and some selected planets

  4. all objects in the universe
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gravity is not a property of the earth alone. Every object in the universe exerts a force on every other object. This force is known as the gravitational force.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance ?

  1. $\dfrac { \pi R }{ 2 } $
  2. $\dfrac { 2R }{ 3 } $
  3. $\dfrac { R }{ 2 } $
  4. $\dfrac { 4R }{ 3\pi } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The centre of mass of a uniform thin hemispherical shell of radius R is located at a distance  $\dfrac{R}{2}$ from the center.

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

The centre of mass of a system of particles is at the origin. It follows that:

  1. the number of particles to the right of the origin is equal to the left of origin.

  2. the total mass of the particles to the right of the origin is same as total mass to the left of the origin.

  3. the number of particles on the X-axis should be equal to the number of particles on the Y-axis .

  4. if there is a particle on the +ve X-axis, there should be atleast one particle on the -ve X-axis.

  5. None of these.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Center of mass of a system of particles is at the origin. It follows that:

(a) Number of particles to the right of origin=Number of particles to the left of origin
(b) Total mass of particles to the right of origin=Total mass of particles to the left of origin

Multiple choice physics turning effects of forces stability and centre of mass center of mass centre of mass

A uniform metal disc of radius R is taken and out of it a disc of diameter $\dfrac{R}{2}$ is cut off from the end.The centre of mass of the remaining part will be :

  1. $\dfrac{R}{28}$ from the centre
  2. $\dfrac{R}{3}$ from the centre
  3. $\dfrac{R}{5}$ form the centre
  4. $\dfrac{R}{6}$ from the centre
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
No correct option.

Given,

$Radius =R,D=\dfrac{R}{2}$

Let origin be the center  

$0=\dfrac{m _1x _1+m _2x _2}{m _1+m _2}$

Mass of the diameter $\dfrac{R}{2}=\dfrac{m}{8}$

$0=\dfrac{\dfrac{7m}{8}X _1+\dfrac{m}{8}(\dfrac{R}{4})}{\dfrac{7m}{8}+\dfrac{m}{8}}=\dfrac{7m}{8}x _1+\dfrac{m}{8}\times\dfrac{R}{4}\Rightarrow x _1=\dfrac{-R}{28}$