Q#8 (Past Exam Paper – June 2016 Paper 42 Q1)
A binary star consists of two stars A and B that orbit one another, as illustrated in Fig. 1.1.
Fig. 1.1
The stars are in circular orbits with the centres of both orbits at point P, a distance d from the centre of star A.
(a) (i) Explain why the centripetal force acting on both stars has the same magnitude. [2]
(ii) The period of the orbit of the stars about point P is 4.0 years.
Calculate the angular speed ω of the stars. [2]
(b) The separation of the centres of the stars is 2.8 × 108 km.
The mass of star A is MA. The mass of star B is MB.
The ratio MA/MB is 3.0.
(i) Determine the distance d. [3]
(ii) Use your answers in (a)(ii) and (b)(i) to determine the mass MB of star B.
Explain your working. [3]
[Total: 10]
Solution:
(a) (i) The gravitational force provides the centripetal force. From Newton’s 3rd law, the forces have the same magnitude.
(ii)
{Period T = 4 years = (4.0 × 365 × 24 × 3600) seconds}
ω = 2π / T
ω = 2π / (4.0 × 365 × 24 × 3600)
ω = 5.0 (4.98) × 10–8 rad s–1
(b)
(i)
{Centripetal force = mrω2
The centripetal forces acting on the stars are equal in magnitude.
Centripetal force on star A = MA d ω2
Centripetal force on star B = MB (2.8×108 – d) ω2 }
(centripetal force =) MA d ω2 = MB (2.8×108 – d) ω2
{ MA d ω2 = MB (2.8×108 – d) ω2
MA d = MB (2.8×108 – d)
MA / MB = (2.8×108 – d) / d
3.0 = (2.8×108 – d) / d }
MA / MB = 3.0 = (2.8×108 – d) / d
{3d = 2.8×108 – d
3d + d = 2.8×108}
d = 7.0 × 107 km
(ii)
{The gravitational force provides the centripetal force.
Note that the separation should be in metres, not kilometres.
We want to find MB. So, we equate the gravitational force to the centripetal force on star A so that MA gets eliminated.
GMAMB / r2 = MAdω2
For the gravitational force, r is the separation of the stars.
For the centripetal force, d is the distance of star A from point P.}
GMAMB / (2.8×1011)2 = MAdω2
MB = (2.8 × 1011)2 × dω2 / G
MB= (2.8×1011)2 × (7.0×1010) × (4.98×10–8)2 / (6.67×10–11)
MB= 2.0 × 1029 kg
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