Q#1 (Past Exam Paper – June 2017 Paper 41 & 43 Q1)
(a) Explain how a satellite may be in a circular orbit around a planet. [2]
(b) The Earth and the Moon may be considered to be uniform spheres that are isolated in space.
The Earth has radius R and mean density ρ. The Moon, mass m, is in a circular orbit about the Earth with radius nR, as illustrated in Fig. 1.1.
Fig. 1.1
The Moon makes one complete orbit of the Earth in time T.
Show that the mean density ρ of the Earth is given by the expression
ρ = 3πn3 / GT2
[4]
(c) The radius R of the Earth is 6.38 × 103 km and the distance between the centre of the Earth and the centre of the Moon is 3.84 × 105 km.
The period T of the orbit of the Moon about the Earth is 27.3 days.
Use the expression in (b) to calculate ρ. [3]
[Total: 9]
Solution:
(a) The gravitational force (of attraction between the satellite and the planet) provides / is the centripetal force (on the satellite about the planet)
(b)
{Density = Mass / Volume and Volume of Earth (sphere) = (4/3) × πR3 }
Mass M = (4/3) × πR3 ρ
{ω = 2 π / T
or speed v = distance / time = circumference / period}
ω = 2π / T
or v = 2πnR / T
{The gravitational force provides the centripetal force.
GMm / r2 = mrω2
or GMm / r2 = m v2 / r
where r is the radius of the orbit
The mass of the Moon m may be eliminated on both sides.
Radius of orbit r = nR
Replacing the equations for mass M and ω or v,}
GM / (nR)2 = nRω2
or v2 / nR
Consider GM / (nR)2 = nRω2
M = nRω2 × (nR)2 / G
M = ω2 × (n3R3 / G)
M = ω2 × (n3R3 / G)
(4/3) × πR3 ρ= (2π / T)2 × (n3R3 / G)
substitution clear to give ρ = 3πn3 / GT2
(c)
{ρ= 3πn3 / GT2 }
{nR = 3.84 × 105 km
and R = 6.38 × 103 km }
n = (3.84 × 105) / (6.38 × 103) = 60.19 or 60.2
{T = 27.3 days = 27.3 × 24 × 3600 s}
ρ = 3π × 60.193 / [(6.67 × 10-11) × (27.3 × 24 × 3600)2]
ρ = 5.54 × 103 kg m-3
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