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 physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A person sitting in a chair in a satellite feels weightless because

  1. the earth does not attract the object in a satellite.

  2. the normal force by the chair on the parson balance the earht's attraction.

  3. the normal force is zero.

  4. the person is satellite is not accelerated.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As a person sits in a chair $($ on Earth's surface $)$, he will experience two forces, the force of the Earth's gravitational field pulling him downward toward the Earth and the force of the chair pushing him upward. The upward chair force is sometimes referred to as a normal force.
If there were no upward normal force acting upon body, body would not have any sensation of the weight. Without the contact force $($ the normal force $)$, there is no means of feeling the non-contact force $($ the force of gravity $)$.
Hence, A person sitting in a chair in a satellite feels weightless because normal force is zero.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

In an earth satellite moving in a circular orbit, a piece of metal weighing $16g$ (on the earth) is weighed by a spring balance while the metal is suspended in water. If the relative density of the metal is $8$, what weight will be recorded?

  1. $-2$ g

  2. zero
  3. $2$ g
  4. $14$ g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In the satellite the spring is in free fall so no weight will be recorded.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

An astronaut, inside an earth satellite, experiences weightlessness because

  1. no external force is acting on him

  2. he is falling freely

  3. no reaction is exerted by the floor of the satellite

  4. he is far away from the earth's surface
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

For an earth satellite moving in a circular orbit, centripetal force required for its circular motion is provided by the gravitational force exerted by earth on it.
 It means resultant force on the astronaut is equal to the gravitational force exerted by earth on him. Hence, no reaction is exerted by floor of the satellite on him.
In other words, his acceleration towards earth centre (centripetal acceleration) is exactly equal to acceleration caused by the gravitational force alone.
Hence, options (b) and (c) are correct.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

Moon is a satellite of the Earth, but weightlessness is not experienced at the surface of the Moon because

  1. its distance from the Earth is more.

  2. it is a natural satellite.

  3. its size is big but density is very low.

  4. its own mass is more.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As Moon has its mass, thus it posses its own gravity ($\frac {1}{6}$th of that of the Earth). Hence, weightlessness is not experienced at the surface of the Moon.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

The International Space Station is currently under construction. Eventually, simulated earth gravity may become a reality on the space station. What would the gravitational field through the central axis be like under these conditions?

  1. Zero
  2. $0.25\ g$
  3. $0.5\ g$
  4. $0.75\ g$
  5. $1\ g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Simulated earth gravity can be realized by rotating the space station about a central axis. This rotation creates centrifugal force on the people inside the space station away from the central axis. Thus, $g={ \omega  }^{ 2 }R $ at the central axis $R=0$. So, gravitational field is zero.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

The rotation of the Earth having radius R about its axis speed upto a value such that a man at latitude angle $60^o$ feels weightless. The duration of the day in such case will be.

  1. $\displaystyle 8\pi\sqrt{\displaystyle\frac{R}{g}}$
  2. $\displaystyle 8\pi\sqrt{\displaystyle \frac{g}{R}}$
  3. $\displaystyle \pi\sqrt{\displaystyle\frac{R}{g}}$
  4. $\displaystyle 4\pi\sqrt{\displaystyle\frac{g}{R}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a man at an angle $\theta=60^o$

$T=2\pi\sqrt{\cfrac{R^3}{GM(\cos60^o)}}$
$2\pi \sqrt{\cfrac{R^2}{GM}\cfrac{R}{(\cos 60^o}}$
$2\pi \sqrt{\cfrac{R}{g}\cfrac{1}{1/2}}$
$=8\pi\sqrt{\cfrac{R}{g}}$

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

The rotation of the Earth having radius $R$ about its axis speeds upto a value such that a man at latitude angle $60^o$ feels weightless. The duration of the day in such case will be.

  1. $8\pi\sqrt{\displaystyle \frac{R}{g}}$
  2. $8\pi\sqrt{\displaystyle \frac{g}{R}}$
  3. $\pi\sqrt{\displaystyle \frac{R}{g}}$
  4. $4\pi\sqrt{\displaystyle \frac{g}{R}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$0=g-{ Rw }^{ 2 }\cos ^{ 2 }{ 60° } \ { w }^{ 2 }=\cfrac { 4g }{ R } \quad or,{ w }^{ 2 }=2\sqrt { \cfrac { g }{ R }  } \ \cfrac { 2\pi  }{ T } =2\sqrt { \cfrac { g }{ R }  } \ \therefore T=\pi \sqrt { \cfrac { R }{ g }  } $

Multiple choice physics pressure in liquids and gases common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

The atmospheric pressure and height of barometer column is $10^5 P _a$ and 760mm respectively on the earth surface. If the barometer is taken to moon then column height will be

  1. zero

  2. 76 mm

  3. 126.6 mm

  4. 760 mm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since there is no atmosphere on moon, no atmospheric pressure will be there and hence corresponding to 0 pressure, height of mercury column will be 0.


Answer is option A.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

Let $F _{1}$ be the magnitude of the gravitational force exerted on the Sun by Earth and $F _{2}$ be the magnitude of the force exerted on Earth by the Sun. Then.

  1. $F _{1}$ is much greater than $F _{2}$
  2. $F _{1}$ is slightly greater than $F _{2}$
  3. $F _{1}$ is equal to $F _{2}$
  4. $F _{1}$ is slightly less than $F _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. Therefore, the gravitational force exerted by the Sun on the Earth is equal in magnitude to the force exerted by the Earth on the Sun.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

The mass of the moon is about $1.2$% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth

  1. Is the same

  2. Is smaller

  3. Is greater

  4. Varies with its phase
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

By Newton's Third Law and Newton's Law of Universal gravitation, the gravitational force the Earth exerts on the Moon is exactly the same as the force the Moon exerts on the Earth. 
F (Earth) = F (Moon)
However, the tidal forces are not equal - the Moon exerts a greater tidal force because tidal forces result from a difference in force across the object. Because the Earth has a greater diameter, the difference in the lunar force across the Earth will be greater than the difference that the terrestrial force exerts across the lunar diameter.

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

The gravitational force between two bodies is

  1. repulsive at large distances

  2. attractive at all places

  3. attractive at short distances

  4. repulsive at short distances
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Gravitation is a natural phenomenon by which all things with mass are brought towards one another. It is an attractive force and behaves uniformly for all values of $r$ (distance between the two bodies).

Multiple choice physics universe and space heliocentric model introduction to gravitation introduction to gravity

On the surface of the earth, force of gravitational attraction between two masses kept at distance d apart is 6 Newtons. If these two masses are taken to the surface of the moon and kept at the same distance d, the force between them will be

  1. 1N

  2. 36N
  3. $\frac { 1 }{ 6 } $N

  4. 6N
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gravitational force by Newton's law must remain the same.


hence, $(D)$ is correct.