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 the measurement of Earth's gravity?

  1. 8.2 m/s2

  2. 3.5 m/s2

  3. 9.8 m/s2

  4. 4.7 m/s2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The standard acceleration due to gravity on Earth is approximately 9.8 m/s^2. The other options are incorrect values.

Multiple choice physics pressure in fluids and atmospheric pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

Atmospheric pressure exerted on earth is due to the.

  1. Gravitational pull

  2. Revolution of earth

  3. Rotation of earth

  4. Uneven heating of earth
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The atmospheric pressure exerted on earth is due to gravitational pull of the earth and is about 14.7 psi.

Hence, option A is correct.

Multiple choice physics pressure in fluids and atmospheric pressure pressure in air introduction to atmospheric pressure pressure exerted by air devices to measure pressure

Which of the following statement is NOT true.

  1. A freely falling body is acted upon by gravitational force

  2. pressure on top of mount everest is much more than 1 atm

  3. S.I. unit of pressure is pascal

  4. the blood vessels in our body maintains the internal pressure equal to the atmospheric pressure
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation

A free falling object is an object that is falling under the sole influence of gravity.
Atmospheric pressure decreases with height above the surface of a planet because there is less total mass in the atmosphere above a reference point as the height of the reference point increases. The standard atmospheric pressure at sea level is 1 atmosphere.
Pressure is generally measured in units of Pascals and S.I. unit of pressure is pascal.
The normal atmospheric pressure is 760 mm of Hg. But the normal human blood pressure is around 120/80 mm only.
Answer (B) & (D) 
(B) pressure on top of mount everest is much more than 1 atm
(D) the blood vessels in our body maintains the internal pressure equal to the atmospheric pressure

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

The distance between the centre of the earth and moon is 384000 km. If the mass of the earth is $6 \times 10^{24} kg$ and $G=6.66\times 10^{-11}$ units,the speed of the moon is nearly

  1. 1 km/s

  2. 4 km/s

  3. 8 km/s

  4. 11.2 km/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Orbital speed v = sqrt(GM/r). G = 6.67 * 10^-11, M = 6 * 10^24, r = 3.84 * 10^8 m. v = sqrt(6.67 * 10^-11 * 6 * 10^24 / 3.84 * 10^8) = sqrt(40.02 * 10^13 / 3.84 * 10^8) = sqrt(10.4 * 10^5) approx 1020 m/s = 1 km/s.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?

  1. $\dfrac{2R}{5}$
  2. $\dfrac{2R}{3}$
  3. $\dfrac{3R}{7}$
  4. $\dfrac{R}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

$\begin{array}{l} \dfrac { { GM } }{ { { { \left( { R+h } \right)  }^{ 2 } } } } =\dfrac { { 36 } }{ { 100 } } \dfrac { { Gm } }{ { { R^{ 2 } } } }  \ \Rightarrow 100{ R^{ 2 } }=36{ \left( { R+h } \right) ^{ 2 } } \ \Rightarrow 25{ R^{ 2 } }=9\left( { { R^{ 2 } }+{ h^{ 2 } }+2Rh } \right)  \ \Rightarrow 25{ R^{ 2 } }=9{ R^{ 2 } }+9{ h^{ 2 } }+18Rh \ \Rightarrow 16{ R^{ 2 } }=9{ h^{ 2 } }+18Rh \ \Rightarrow 9{ h^{ 2 } }+18Rh-16{ R^{ 2 } }=0 \ R=\dfrac { { -18+\sqrt { 324+576 }  } }{ { 18 } } R \ =\dfrac { { -18+30 } }{ { 18 } } R \ =\dfrac { { 12R } }{ { 18 } } =\dfrac { { 2R } }{ 3 }  \ h=\dfrac { { 2R } }{ 3 }  \end{array}$
Then,
Option $B$ is correct answer.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

An object weighs 10$\mathrm { N }$ at the north pole of the Earth. In a geostationary satelite at a distance of 7R from the centre of the Earth (of radius $\mathrm { R } )$ , the true weight and the apparent weight are respectively.-

  1. 0,0
  2. $0.2 \mathrm { N } , 0$
  3. $0.2 \mathrm { N } , 9.8 \mathrm { N }$
  4. $0.2 N , 0.2 \mathrm { N }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

An object is released from rest at a distance of ${2r} _{e}$ from the center of the Earth, where ${r} _{e}$ is the radius of the Earth. Find out the velocity of the object when it hits the Earth in terms of the gravitational constant $\left(G\right)$, the mass of the Earth $\left(M\right)$, and ${r} _{e}$.

  1. $\sqrt{{GM}/{{r} _{e}}}$
  2. ${GM}/{{r} _{e}}$
  3. $\sqrt{{GM}/{2{r} _{e}}}$
  4. ${GM}/{2{r} _{e}}$
  5. $2{GM}/{{r} _{e}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:

Initial velocity, $u=0$  ,  $h=2r _{e}$
Now, by using: $v^{2}=u^{2}+2gh$

                          $v^{2}=0+2g\times2r _{e}$

By putting  $g=GM/{r _{e}}^{2}$ in above equation 
We get, $v^{2}=2GM/{r _{e}}^{2}\times2r _{e}$
             $v^{2}=GM/2r _{e}$
             $v=\sqrt{GM/2r _{e}}$

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

Given that the universal gravitational constant, $G = 6.7 10^{-11} Nm^{2} kg^{-2}$ and that the mass,
M of the earth is $6.0 10^{24} kg$, find the speed of a satellite that is fixed to permanently
focus on the city of Abuja for broadcast of the 2010 IJSO competition.

  1. $ 3.08 \times 10^{3} ms^{-1}$
  2. $24 ms^{-1}$
  3. $40 ms^{-1}$
  4. $3.66 10^{3} ms^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Centripetal force $=$ gravitational force
,$ \Rightarrow   mv _{2}r=GmMr _{2} , from     where    v _{2}=GMr$
where v is the velocity. Period, $T=2\pi rv$ giving
$r=Tv2\pi$
Hence $v _{3}=2\pi GMT   With   T = 24 hr = 8.64 \times  104 s$
$v=32\pi\times 6.7 10^{-11} \times 6.0 \times 10248.64\times 104 = 3.08 10^{3} ms$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A body of density ${d _1}$ is counterpoised by $Mg$ of weights of density ${d _2}$ in air of density $d.$ Then the true mass of the body is

  1. $M$
  2. $M\left( {1-\dfrac{d}{{{d _2}}}} \right)$
  3. $M\left( {1 - \dfrac{d}{{{d _1}}}} \right)$
  4. $\dfrac{{M\left( {1 - d/{d _1}} \right)}}{{\left( {1 - d/{d _2}} \right)}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The true mass M_true is found by equating the forces: M_true * g - V * d * g = M * g - V * d2 * g, where V is volume. Substituting V = M_true / d1 leads to the correct buoyant force balance equation.

Multiple choice biology movement and locomotion in living organisms movements in plants coordination in plants tropic and nastic movements in plants

Geotropism is

  1. Growth away from the vector of gravity

  2. Growth at right angles to the force of gravity

  3. Response to the stimulus of gravity

  4. Unequal growth due to gravity
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Geotropism is the plant response to the stimulus of gravity. Plant growth away from the gravity is called as negative geotropism. Stem shows negative geotropism. Plant growth towards the gravity is called as positive geotropism. Root shows positive geotropism.

Thus, the correct answer is option C.

Multiple choice evs - i our earth and our solar system solar system and sun planets and stars planets sunita's experience in space space exploration

If earth suddenly stops rotating, then the weight of an object of mass $m$ at equator will [$\omega$ is angular speed of earth and $R$ is its radius ] :

  1. Decrease by $m\omega^2R$
  2. Increase by $ m\omega^2R$
  3. Decrease by $m\omega R^2$
  4. Increase by $ m\omega R^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Weight when the earth rotates is
$m{g}^{\prime}=mg-mR{\omega}^{2}$
When $\omega=0$
The weight will increase by $mR{\omega}^{2}$
Multiple choice physics wave motion wave velocity speed and acceleration of travelling wave speed of a travelling wave

Common balance reading on earth of an object is 'X' and the spring balance reading is 'Y'. The set up is transferred to moon. What reading does the common balance and spring balance give?

  1. $\dfrac{X}{6}, Y$
  2. $X, \dfrac{Y}{2}$
  3. $X, \dfrac{Y}{6}$
  4. $X, \dfrac{Y}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A common balance (beam balance) measures mass, which is invariant under gravity. A spring balance measures weight (W = mg), which depends on the local acceleration due to gravity g. On the moon, g is 1/6 of that on Earth, so the spring balance reading becomes Y/6.

Multiple choice

If speed of rotation of the earth increases, weight of the body

  1. increases

  2. remains unchanged

  3. decreases

  4. . may decrease or increase
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The Earth's rotation creates a centrifugal force that acts outward, opposing gravity. If the speed of rotation increases, this outward force increases, effectively reducing the net gravitational pull experienced by a body, thus decreasing its weight.