Gravitation

The mass of a planet is $\frac{1}{10}$ th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

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Contributor-Level 10

$g^{'} = \frac{G M^{'}}{R^{' 2}} = \frac{G M}{10 {(\frac{R}{2})}^{2}}$

$= \frac{4}{10} \frac{G M}{R^{2}} = 0.4 * 9.8$

$= 3.92 m s^{- 2}$

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6 months ago

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to earth reach height h is________

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Contributor-Level 10

Using conservation of energy :

$? \frac{? G M m}{R} + \frac{1}{2} m {(\sqrt[]{\frac{2 G M}{R}})}^{2} = \frac{? G M m}{r} + \frac{1}{2} m v^{2}$

$? \frac{1}{2} m v^{2} = \frac{G M m}{r}$

$? ?_{R}^{R + h} r^{1 / 2} d r = \sqrt[]{2 G M} ?_{0}^{t} d t$

$? \frac{2}{3} R^{3 / 2} [{(1 + \frac{h}{R})}^{3 / 2} ? 1] = \sqrt[]{2 G M} t$

$t = \frac{1}{3} \sqrt[]{\frac{2 R}{g}} [{(1 + \frac{h}{R})}^{3 / 2} ? 1]$

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Consider a binary star system of star A and star B with masses m_A and m_B revolving in a circular orbit of radii r_A and r_B, respectively. If T_A and T_B are the time period of star A and star B, respectively, then:

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Since binary mass system performs circular motion about is common centre of mass, so

$m_{A} ω_{A}^{2} r_{A} = \frac{G m_{B} m_{A}}{{(r_{A} + r_{B})}^{2}} = \frac{G m_{B} m_{A}}{r^{2}}$

$m_{A} ω_{A}^{2} * \frac{m_{B}}{(m_{A} + m_{B})} r = \frac{G m_{B} m_{A}}{r^{2}}$

$ω_{A} = \sqrt[]{\frac{G (m_{A} + m_{B})}{r^{3}}}$

Similarly we can show that

$ω_{B} = \sqrt[]{\frac{G (m_{A} + m_{B})}{r^{3}}}$

Hence their angular velocity will be same, time period will be same, i.e. TA = TB

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Four particle each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

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Contributor-Level 10

For circular motion F_{net

$= \frac{M V^{2}}{R}$}

_{$\sqrt[]{2 F} + F' = \frac{M V^{2}}{R}$}

$\sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} R)}^{2}} + \frac{G M M}{{(2 R)}^{2}} = \frac{M V^{2}}{R}$

$\frac{G M}{R} [\frac{1}{\sqrt[]{2}} + \frac{1}{4}] = V^{2}$

$V = \frac{1}{2} \sqrt[]{\frac{G M}{R} (2 \sqrt[]{2} + 1)}$

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Two satellites A and B, having masses in the ration 4 : 3, are revolving in circular orbits of radii 3r and 4r respectively around the earth. The ratio of total mechanical energy of A to B is:

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Contributor-Level 10

$\frac{E_{A}}{E_{B}} = \frac{- \frac{G M m_{A}}{2 * 3 r}}{\frac{- G M m_{B}}{2 * 4 r}} = \frac{m_{A}}{m_{B}} * \frac{4}{3} = \frac{4}{3} * \frac{4}{3} = \frac{16}{9}$

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A box weighs $196 N$ on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take $g = 10 {m s}^{- 2}$ at the north pole and the radius of the earth $= 6400 k m$ )

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Contributor-Level 10

$W_{equator} = W_{pole} - m ω^{2} R$

$\begin{array}{r} = 196 - 19.6 * {(\frac{2 π}{24 * 60 * 60})}^{2} * 6400 * 10^{3} \\ = 195.33 N \end{array}$

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The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is :

[Radius of earth R = 6400 km and $\sqrt[]{3} = 1.732]$

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Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Contributor-Level 10

$g = \frac{9}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow \frac{g}{3} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow {(1 + \frac{h}{R})}^{2} = 3$

1 + $\frac{h}{R} = \sqrt[]{3}$ $\frac{}{}$

$\Rightarrow \frac{h}{R} = \sqrt[]{3} - 1$

$\frac{h}{R} = 1.732 - 1$

$\frac{h}{R} = 0.732$

h = 0.732 * 6400

= 4684.8 km » 4685 km

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6 months ago

Two objects of equal masses placed at certain from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other, then the new force will be:

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Gravitation physics ncert exemplar solutions class 12th chapter two

Contributor-Level 10

F = $\frac{G m_{2}}{d^{2}}$ - (i)

Now, $F' = \frac{G m_{1} m_{2}}{d^{2}} = \frac{G 2 m}{3} * \frac{4 m}{3 d^{2}}$

$k' = \frac{8}{9} G \frac{m^{2}}{d^{2}}$

$∴ F' = \frac{8}{G} F$

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Given below are two statements : One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A : If we move form poles to equator, the direction of acceleration due to gravity of earth always points towards the centre of earth without any variation in its magnitude.

Reason R : At equator, the direction of acceleration due to the gravity is towards the centre of earth.

In the light of above statements, chose the correct answer from the options given below

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Gravitation Physics NCERT Exemplar Solutions Class 12th Chapter Eight

Contributor-Level 10

$g' = g - ω^{2} R s i n θ$

$g' (a t θ = 0 p o l e) = g$

$g' (a t θ = 90 °, e q u a t o r) = g - ω^{2} R$

g decreases as we move bole to equator

So, A is false statement

But 'R' statement is true

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The escape velocity of a body on a planet 'A' is 12 kms^-1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is:

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