The time period of a satellite of earth is 5, | vedclass

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Question

$\mathrm{T} \propto \mathrm{r}^{\frac{3}{2}}$

$\frac{\mathrm{T}^{1}}{\mathrm{T}}=(4)^{\frac{3}{2}}$

$\mathrm{T}^{1}=5 	imes 8=40$

Vedclass · Accepted Answer

$40$

Vedclass · Answer

$\mathrm{T} \propto \mathrm{r}^{\frac{3}{2}}$

$\frac{\mathrm{T}^{1}}{\mathrm{T}}=(4)^{\frac{3}{2}}$

$\mathrm{T}^{1}=5 	imes 8=40$

$(a)$ Gravitational constant $(G)$	$(i)$ $\left[ L ^{2} T ^{-2}\right]$
$(b)$ Gravitational potential energy	$(ii)$ $\left[ M ^{-1} L ^{3} T ^{-2}\right]$
$(c)$ Gravitational potential	$(iii)$ $\left[ LT ^{-2}\right]$
$(d)$ Gravitational intensity	$(iv)$ $\left[ ML ^{2} T ^{-2}\right]$

The time period of a satellite of earth is $5\, hours$. If the separation between the centre of earth and the satellite is increased to $4\, times$ the previous value, the new time period will become ....... $h$