Physics Gravitation

A body of mass 60g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is :

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Raj Pandey

Contributor-Level 9

$I_{g} = \frac{F}{m} = \frac{3}{60 * 10^{- 3}} = 50 N / k g$

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2 months ago

A satellite is orbiting just above the surface of the earth with period T. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\frac{3 π}{G d}$ represents:

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Physics Gravitation Gravitation

Contributor-Level 10

${\overset{?}{B}}_{P} = {\overset{?}{B}}_{upper wire} \otimes + {\overset{?}{B}}_{semi-circle} ? + {\overset{?}{B}}_{lower-wire} \otimes$

$B_{P} = - \frac{μ_{0} i}{4 π R} + \frac{μ_{0} i}{4 R} - \frac{μ_{0} i}{4 π R}$ $= \frac{μ_{0} i}{4 R} (1 - \frac{2}{π})$ pointing away from the page

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Two bodies of mass $m$ and $9 m$ are placed at a distance $R$ . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( $G =$ gravitational constant):

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Contributor-Level 10

Stress $= \frac{F}{A} = \frac{T}{A} = \frac{W}{A}$

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If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately :
[Take g = 10 ms⁻², the radius of earth, R = 6400*10³ m, Take π = 3.14]

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Contributor-Level 10

g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)

⇒ T = 2π/ω = 2π√ (R/g) = 2 * 3.14 * √ (6400 * 10³)/10)

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The angular momentum of a planet of mass M moving around the sun in an elliptical orbit is L. The magnitude of the areal velocity of the planet is:

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Contributor-Level 10

A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)

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A particle of mass m moves in a circular orbit under the central potential field, V(r) = -C/r, where C is a constant. The correct radius-velocity graph of the particle's motion is;

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Contributor-Level 10

U = mV (r) = -Cm/r

F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force

⇒ mv²/r = Cm/r² ⇒ r = C/v²

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A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is ---------. 'P' has the time period of 24 hours.

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Contributor-Level 10

(T / 24) = (3/12)^ (3/2) ⇒ T = 3 hours

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The time period of satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is:

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Raj Pandey

Contributor-Level 9

T² ∝ R³
⇒ (T? /T? )² = (R? /R? )³ ⇒ T? = 9³/² T = 27T

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2 months ago

The radius in kilometre to which the radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is .

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Contributor-Level 10

The formula for escape velocity (v_e) is v_e = √ (2GM/R).
According to the question, the new escape velocity (v_e') from a new radius R' is related to the original escape velocity by 10v_e' = v_e.
10 * √ (2GM/R') = √ (2GM/R)
Squaring both sides:
100 * (2GM/R') = (2GM/R)
100/R' = 1/R
R' = R/100

If R is the radius of Earth (6400 km), then:
R' = 6400 km / 100 = 64 km

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The value of the acceleration due to gravity is g₁ at a height h = R/2 (R = radius of the earth) from the surface of the earth. It is again equal to g₁ at a depth d below the surface of the earth. The ratio (d/R) equals:

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