Q#4 (Past Exam Paper – November 2015 Paper 41 &42 Q1)
A satellite of mass mS is in a circular orbit of radius x about the Earth.
The Earth may be considered to be an isolated uniform sphere with its mass M concentrated at its centre.
(a) (i) Show that the kinetic energy EK of the satellite is given by the expression
EK = GMmS / 2x
where G is the gravitational constant. Explain your working. [3]
(ii) State an expression, in terms of G, M, mS and x, for the potential energy EP of the
satellite. [1]
(iii) Using answers from (i) and (ii), derive an expression for the total energy ET of the
satellite. [2]
(b) Small resistive forces acting on the satellite cause the radius of its circular orbit to change.
Use your answers in (a) to state, for the satellite, whether each of the following quantities
increases, decreases or remains constant.
(i) total energy [1]
(ii) radius of orbit [1]
(iii) potential energy [1]
(iv) kinetic energy [1]
Solution:
The gravitational force provides the centripetal force.
{Gravitational force = Centripetal force}
GMmS / x2 = mSv2 / x
{msv2 = GMms / x
Multiply by half on both sides,
½ msv2 = GMms / 2x
EK = GMms / 2x}
(ii) EP = – GMmS / x
(iii)
ET = EK + EP
ET = GMmS / 2x – GMmS / x
ET = – GMmS / 2x
(b)
(i) decreases
{The total energy decreases as work needs to be done against the resistive forces.}
(ii) decreases
{Since the total energy decreases, the radius of orbit also decreases.
ET = – GMmS / 2x
The total energy decreases, that is, it becomes more negative. The value of (GMmS / 2x) should be greater so that the total energy (with its negative sign) decreases. Since the value of (GMmS / 2x) increases, the radius of orbit, x, decreases.}
(iii) decreases
{The smaller the radius of orbit, the lower the potential energy. Recall that the potential energy is maximum (and equal to zero) at infinity.}
(iv) increases
{EK = GMms / 2x
From the formula, as the radius x decreases, the EK increases.}
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