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YOUR PARTICIPATION FOR THE GROWTH OF PHYSICS REFERENCE BLOG

Monday, December 28, 2020

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]

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