Question #0091d

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Question #0091d

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Answer 1 · 2016-12-01T05:19:33

Let

$m \to mass of satellite$ $m->"mass of satellite"$

$M \to mass of Earth$ $M->"mass of Earth"$

$R \to radius of the orbit of the satellite$ $R->"radius of the orbit of the satellite"$

$G \to Gravitational constant$ $G->"Gravitational constant"$

$T \to time period of satellite$ $T->"time period of satellite"$

$ω \to angular speed of satellite = \frac{2 π}{T}$ $omega->"angular speed of satellite"=(2pi)/T$

The centripetal force ( $F_{c}$ $F_c$ ) acting on the satellite revolving round the Earth along the orbit having radius R is given by

$F_{c} = m ω^{2} R$ $F_c=momega^2R$

The gravitaional force $(F_{g})$ $(F_g)$ acting on the satellite will provide the required centripetal force.

$F_{g} = G \frac{m M}{R^{2}}$ $F_g=G(mM)/R^2$

Now $F_{c} = F_{g}$ $F_c=F_g$

$\Rightarrow m ω^{2} R = \frac{G m M}{R^{2}}$ $=>momega^2R=(GmM)/R^2$

$\Rightarrow {(\frac{2 π}{T})}^{2} R = \frac{G M}{R^{2}}$ $=>((2pi)/T)^2R=(GM)/R^2$

$=>T^2=((4pi^2)/(GM))R^3....... (1)$

Now differentiating (1) w.r to R we get

$=>2TdT=((4pi^2)/(GM))*3R^2dR........(2)$

Dividing (2) by (1) we get

$2(dT)/T=(3dR)/R$

$=>(dT)/T=3/2((dR)/R)......(3)$

Now by the problem change in radius of the satellite is
$dR=1.02R-R=0.02R$

Inserting this value of $dR=0.02R$ in equation (3) we get the percentage change in time period of second satellite w.r to the time period of first satellite.

$"% change in time period"$

$=(dT)/Txx100%=3/2((0.02R)/R)xx100%$

$=3%$

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Question #0091d

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