Physics Gravitation

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New answer posted•a year ago

Physics Gravitation Class 11th

Suppose two planets (spherical in shape) of radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is √(aGM/(7R)) then the value of 'a' is _______. [Given: The two planets are fixed in their position]

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

We have to find the point where the gravitational field must be zero.
EG = 0
GM/x² = G (9M)/ (8R-x)²
1/x² = 9/ (8R-x)²
8R - x = 3x => x = 2R
Potential at A (surface of first planet), VA = -GM/R - G (9M)/7R = -16GM/7R
Potential at point x, Vx = -GM/x - G (9M)/ (8R-x) = -GM/2R - G (9M)/6R = -2GM/R
ΔV = Vx -

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New answer posted•a year ago

Physics Gravitation Class 11th

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

g = (4/3)πRρG . (i) (ρ = density)
Now ρ = M/V = M/ (4/3)πR³)
R³ = M/ (4/3)πρ) => R ∝ (M/ρ)¹/³
From equation (i)
g ∝ Rρ ∝ (M/ρ)¹/³ρ = M¹/³ρ²/³
For planet, M' = 2M, ρ' = ρ
g'/g = (M'/M)¹/³ (ρ'/ρ)²/³ = (2)¹/³ (1)²/³ = 2¹/³
W' = mg' = m (2¹/³g) = 2¹/³ (mg) = 2¹/³W

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New answer posted•a year ago

Physics Gravitation Class 11th

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$m ω^{2} . \frac{2 d}{3} = \frac{G . m . 2 m}{d^{2}} \Rightarrow ω = \sqrt[]{\frac{3 G m}{d^{3}}}$

$T = \frac{2 π}{ω} = 2 π \sqrt[]{\frac{d^{3}}{3 G m}}$

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New answer posted•a year ago

Physics Gravitation Class 11th

Four identical particles of equal masses 1kg made to move along the circumference of a circle of radius 1m under the actin of their own mutual gravitational attraction. The speed of each particle will be:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$F_{1} c o s 45 ° + F_{2} c o s 45 ° + F_{3} = \frac{m v^{2}}{r}$

$\Rightarrow \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(\sqrt[]{2} r)}^{2}} . \frac{1}{\sqrt[]{2}} + \frac{G m^{2}}{{(2 r)}^{2}} = \frac{m v^{2}}{r}$

$\Rightarrow v = \sqrt[]{\frac{G m}{4 r} (2 \sqrt[]{2} + 1)}$

Putting m = 1 kg and r = 1 m,

$v = \frac{1}{2} \sqrt[]{G (1 + 2 \sqrt[]{2})}$

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New answer posted•a year ago

Physics Gravitation Class 11th

Consider two satellites S₁ and S₂ with periods of revolution 1hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S₁ to the angular velocity of satellites S₂ is:

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$ω = \frac{2 π}{T} \Rightarrow \frac{ω_{1}}{ω_{2}} = \frac{T_{2}}{T_{1}} = 8$

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New answer posted•a year ago

Physics Gravitation Class 11th

In the reported figure of earth, the value of acceleration due to gravity is same at point A and C but it is smaller than that of its value at point B (surface of the earth). The value of OA: AB will be x : y. The value of x is…………

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{G M}{{(3 R / 2)}^{2}} = \frac{G M}{R^{3}} \times r$

$\Rightarrow O A = \frac{4 R}{9} = r$

$A B = R - \frac{4 R}{9} = \frac{5 R}{9} \Rightarrow O A : A B = 4 : 5 = x : y \Rightarrow x = 4$

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New answer posted•a year ago

Physics Gravitation Class 11th

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth’s centre, where ‘R’ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

13 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

F = m E cos θ

$= m (\frac{G M}{R^{3}} \times r) \times \frac{x}{r}$ $m a = \frac{m g}{R} \times \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

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New answer posted•a year ago

Physics Gravitation Class 11th

Find the gravitational force of attraction between the ring and sphere as shown in the diagram where the plane of the ring is perpendicular to the joining the centres. If $\sqrt[]{8} R$  is the distance between the centres of a ring (of mass ‘m’) and a sphere (mass ‘M’) where both have equal radius ‘R’.

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Method – I:

$F = M_{s p h e r e} E_{r i n g} = M \times \frac{G m x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = M \times \frac{G m x \sqrt[]{8} R}{{(R^{2} + 8 R^{2})}^{\frac{3}{2}}} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

Method – II:

$F = \int d F c o s θ = c o s θ \int (d m) \frac{G M}{(R^{2} + 8 R^{2})} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

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New answer posted•a year ago

Physics Gravitation Class 11th

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator?

[use g = $\frac{G m}{R^{2}} = 9.8 m s^{- 2}$  and radius of earth, R = 6400km].

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

As we go pole to equator, acceleration due to gravity decreases. So, weight of the body will also reduces. Thus, weight on equator will be slightly smaller than 49 N.

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New answer posted•a year ago

Physics Gravitation Class 11th