Physics Gravitation
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a month agoNew answer posted
a month agoContributor-Level 10
Transverse velocity of rocket
Radial velocity of rocket
Kinetic energy
New answer posted
a month agoContributor-Level 10
F = Gm²/ (2R)² = mω²R Given m = 1kg
⇒ ω² = Gm/ (4R³)
⇒ ω = (1/2)√ (G/R³)
New answer posted
a month agoContributor-Level 10
T² = (4π²r³)/GM
⇒ M = (4π²r³)/ (G T²)
⇒ M = 6 * 10¹¹ * (9 * 10? )³ / (450 * 60)²
= 6.48 x 10²³ kg
(Note: There is a calculation error in the source image. Using the formula and given values: M = (6e11 * (9e6)^3) / (27000)^2 = 6e23 kg)
New answer posted
a month agoContributor-Level 10
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 - VA = -2GM/R - (-16GM/7R) = (-14+16)GM/7R = 2GM/7R
Using conservation of mechanical energy
ΔKE = ΔU = mΔV
½mv² = m (2GM/7R)
v² = 4GM/7R
v = √ (4GM/7R) => a = 4
New answer posted
a month agoContributor-Level 10
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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