Physics Gravitation

a year ago

$T = 2 π \sqrt[]{\frac{r^{3}}{G M_{e}}}$

$T_{B} - T_{A} = \frac{2 π}{\sqrt[]{G M_{e}}} (r_{B}^{3 / 2} - r_{A}^{3 / 2}) = \frac{2 \times 3.14}{\sqrt[]{6.67 \times 1 0^{- 11} \times 6 \times 1 0^{24}}} [{(8 \times 1 0^{6})}^{\frac{3}{2}} - {(7 \times 1 0^{6})}^{\frac{3}{2}}]$

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New answer posted•a year ago

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

Assertion A : The escape velocities of planet Ad and B are same. But A and B are of unequal mass.

Reason R : The product of their mass and radius must be same. M₁ R₁ = M₂R₂

In the light of the above statements, choose the most appropriate answer from the options given below:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

a year ago

${(v_{e})}_{A} = {(v_{e})}_{B} \Rightarrow \frac{M_{1}}{R_{1}} = \frac{M_{2}}{R_{2}}$

R is not correct.

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New answer posted•a year ago

A solid sphere of radius R gravitationally attracts a particle placed at 3R form its centre with a force F₁,. Now a spherical cavity of radius $(\frac{R}{2})$  is made in the sphere (as shown in figure) and the force becomes F₂. The value of F₁: F₂ is:

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Asked by Vishal Baghel

Answered by

Physics Gravitation Class 11th

a year ago

$F_{1} = \frac{G M m}{9 R^{2}}$

$F_{2} = \frac{G M m}{9 R^{2}} - \frac{G \frac{M}{8} m}{{(\frac{5 R}{2})}^{2}} = \frac{G M m}{R^{2}} (\frac{1}{9} - \frac{1}{50})$

$\frac{F_{1}}{F_{2}} = \frac{1}{1 - \frac{9}{50}} = \frac{50}{41}$

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New answer posted•a year ago

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity v_e such that $v_{i} = \sqrt[]{\frac{x}{y}} \times v_{e} .$ The value of x will be………

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Using conservation of energy :

$\frac{1}{2} m v_{i}^{2} - \frac{G m m}{R} = 0 - \frac{G m m}{11 R}$

$v_{i}^{2} = \frac{20}{11} \frac{G m}{R}$

Escape velocity is defined as $\sqrt[]{\frac{2 G M}{R}}$

$v_{i} = \sqrt[]{\frac{10}{11} v_{e}}$

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New answer posted•a year ago

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

a year ago

Apply energy conservation,

$U_{i} + K_{i} = U_{f} + K_{f}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} m v^{2}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} \times m \times \frac{G M}{3 R}$

$\Rightarrow K_{i} = - \frac{1}{6} \frac{G M m}{R} + \frac{G M m}{R}$

$K_{i} = \frac{5}{6} \frac{G M m}{R}$

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New answer posted•a year ago

A parallel plate capacitor is charged by connecting it to a battery through a resistor. If $I$ is the current in the circuit, then in the gap between the plates:

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Asked by Alok Kumar Singh

Answered by

a year ago

According to modified Ampere's law

$? B . d I = μ_{0} (I_{C} + I_{D})$

For Loop $L_{1} I_{C} \neq 0$ and $I_{D} = 0$

For Loop $L_{2} I_{C} = 0$ and $I_{D} \neq 0$

Due to $KCL I_{C} = I_{D}$

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New answer posted•a year ago

The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 × 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Physics Gravitation Class 11th

a year ago

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 × 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π \times 1 0^{3}}{6 \times 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

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New answer posted•a year ago

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be: (Radius of earth will, be

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$g^{1} = g (1 - \frac{2 h}{R}) = g (1 - 2 \times \frac{32}{6400})$

$g^{1} = \frac{99 g}{100} = 0.99 g$

% decrease is wt = $\frac{g - g^{1}}{g} \times 100 = 1 %$

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New answer posted•a year ago

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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Asked by Alok Kumar Singh

Answered by

a year ago

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

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

$= 3.92 m s^{- 2}$

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New answer posted•a year 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________

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by