Explain how a satellite may be in a circular orbit around a planet. The Earth and the Moon may be considered to be uniform spheres that are isolated in space.

Question 1

(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]

Reference: Past Exam Paper – June 2017 Paper 41 & 43 Q1

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