Two small satellites move in circular orbits around the earth- the distances r and r - -r from the centre of the earth- Their time period of rotation are T and T - -T- -r - r- -T - T

The correct option is A Δ T = 3/2 T Δ r/r As T^2∝ r^3⇒ T^2=cr^3 (1) , for some constant c .So, (T+Δ T)^2=c(r+Δ r)^3 .As, Δ T lt;T;Δ ...

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Standard XII

Physics

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Two small sat...

Question

Two small satellites move in circular orbits around the earth, the distances r and $r + Δ$ r from the centre of the earth. Their time period of rotation are $T$ and $T +$ $Δ$ T. ( $Δ$ r << r, $Δ$ T << T)

Δ T = \frac{3}{2} T \frac{Δ r}{r}

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Δ T = - \frac{3}{2} T \frac{Δ r}{r}

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Δ T = \frac{2}{3} T \frac{Δ r}{r}

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Δ T = T \frac{Δ r}{r}

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Solution

The correct option is A $Δ T = \frac{3}{2} T \frac{Δ r}{r}$
As $T^{2} \propto r^{3} \Rightarrow T^{2} = c r^{3} - (1)$ , for some constant $c$ .
So, $(T + Δ T)^{2} = c (r + Δ r)^{3}$ .
As, $Δ T < T; Δ r < r$
So, $T^{2} + 2 T Δ T = c (r^{3} + 3 r^{2} Δ r) - (2)$
From $(1) a n d (2)$
$2 T Δ T = c 3 r^{2} Δ r$
Dividing by $(1)$ ,
$2 Δ T / T = 3 Δ r / r$
$\Rightarrow Δ T = 3 T Δ r / 2 r$

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Two small satellites move in circular orbits around the earth, the distances r and r+Δr from the centre of the earth. Their time period of rotation are T and T+ΔT. (Δr << r, ΔT << T)

The correct option is A ΔT=32TΔrrAs T2∝r3⇒T2=cr3 −(1), for some constant c.So, (T+ΔT)2=c(r+Δr)3.As, ΔT<T;Δr<rSo, T2+2TΔT=c(r3+3r2Δr) −(2) From (1) and (2)2TΔT=c3r2Δr Dividing by (1),2ΔT/T=3Δr/r⇒ΔT=3TΔr/2r

Two small satellites move in circular orbits around the earth, the distances r and $r + Δ$ r from the centre of the earth. Their time period of rotation are $T$ and $T +$ $Δ$ T. ( $Δ$ r << r, $Δ$ T << T)