The planet Jupiter and one of its moons, Io, may be considered to be uniform spheres that are isolated in space.

 Question 11

(a) State Newton’s law of gravitation. [2]

 

 

(b) The planet Jupiter and one of its moons, Io, may be considered to be uniform spheres that are isolated in space.

Jupiter has radius R and mean density ρ.

Io has mass m and is in a circular orbit about Jupiter with radius nR, as illustrated in Fig. 1.1.

 

Fig. 1.1

 

The time for Io to complete one orbit of Jupiter is T.

Show that the time T is related to the mean density ρ of Jupiter by the expression

ρT² = 3πn³/ G

where G is the gravitational constant. [4]

 

 

(c) (i) The radius R of Jupiter is 7.15 × 10⁴ km and the distance between the centres of Jupiter and Io is 4.32 × 10⁵ km.

The period T of the orbit of Io is 42.5 hours.

Calculate the mean density ρ of Jupiter. [3]

 

(ii) The Earth has a mean density of 5.5 × 10³ kg m^-3. It is said to be a planet made of rock.

By reference to your answer in (i), comment on the possible composition of Jupiter. [1]

 

[Total: 10]

 

 

 

Reference: Past Exam Paper – November 2017 Paper 42 Q1

 

 

 

Solution:

(a) Newton’s law of gravitation states that the gravitational force between masses is proportional to the product of the point masses and inversely proportional to the square of their separation.

 

 

(b)

{Density = mass / volume

Mass = Density × Volume

Since Jupiter is considered to be a uniform sphere, its volume = 4/3 πr³}

 

mass of Jupiter (M) = (4 / 3) πR³ ρ

 

{For the centripetal force, we may either use angular velocity or velocity.

Speed = distance / time = circumference / period}

 

ω = 2π / T                               or                    v = 2πnR / T  

 

{The gravitational force provides the centripetal force.

Centripetal force = gravitational force}

(m)ω²x = GM(m) / x²                  or                    (m)v² / x = GM(m) / x²               

 

{Note that        x = radius of orbit = nR

the mass of Jupiter is given in terms of R, its radius}

 

substitution and correct algebra leading to ρT2 = 3πn3 / G                

(m)ω²x = GM(m) / x² 

(2π / T)² nR = G ×((4 / 3) πR³ ρ) / n²R²

(2π / T)² n³R³ = G × (4 / 3) πR³ ρ

ρT² = 3πn³ / G

 

 

(c) (i)

{ ρT² = 3πn³ / G          We need to find n first.}

{R = 7.15 × 10⁴ km     and nR = 4.32 × 10⁵ km                     so, n = nR / R}

 

n = (4.32 × 10⁵) / (7.15 × 10⁴)            or        n = 6.04          

 

ρ × (42.5 × 3600)² = (3π × 6.04³) / (6.67 × 10^-11)                 

ρ = 1.33 × 10³ kg m^-3                                                 

 

 

(ii)

{The density is approximately one fifth of the density of the Earth.}

Jupiter likely to be a gas/liquid (at high pressure)       [allow other sensible suggestions]