Physics Gravitation
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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]
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]
g? = g - ω²R ⇒ 0 = g - ω²R ⇒ ω = √ (g/R)
⇒ T = 2π/ω = 2π√ (R/g) = 2 × 3.14 × √ (6400 × 10³)/10)
Upvote 
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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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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A = Area swept ⇒ dA/dt = (1/2)r² (dθ/dt) = (1/2) (Mr²ω)/M = L / (2M)
Upvote 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;
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;
U = mV (r) = -Cm/r
F = -dU/dr = -Cm/r² ⇒ The force which provides required centripetal force
⇒ mv²/r = Cm/r² ⇒ r = C/v²
Upvote 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.
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.
(T / 24) = (3/12)^ (3/2) ⇒ T = 3 hours
Upvote 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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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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T² ∝ R³
⇒ (T? /T? )² = (R? /R? )³ ⇒ T? = 9³/² T = 27T
Upvote 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 .
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 .
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
Upvote 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:
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:
dW? = Eqdx
∫ dU? = ∫ kQ/x² dxq
W? = kQq (-1/x)|? ^ (R+y)
W? = kQq (y)/ (R) (R + y)
W? + W? = 1/2 mv²
V² = 2/m (kQqy/ (R) (R+y) + mgy)
V² = 2y (kQq/ (m (R) (R+y) + g) ; k = 1/ (4πε? )
Upvote