Question
If the escape velocity of a satellite at a height above the earth's surface is $v=\sqrt{\frac{2 G M}{R}}$. Then check the correctness of the formula.
✓
Answer
It is given, $\quad v=\sqrt{\frac{2 GM }{ R }}$
Dimension of velocity $(v)$ in left side $=\left[ M ^0 L^1 T^{-1}\right]$
Dimension of right side $=\sqrt{\frac{2 GM }{ R }}$
$=\sqrt{\frac{\left[ M ^{-1} L^3 T^{-2}\right][ M ]}{[ L ]}}$
$=\sqrt{\left[ M ^0 L^2 T^{-2}\right]}=\left[ M ^0 L^{ l } T ^{-1}\right]$
$\therefore$ Dimension of left side $=$ Dimension of right side
So, the given formula is correct.
Dimension of velocity $(v)$ in left side $=\left[ M ^0 L^1 T^{-1}\right]$
Dimension of right side $=\sqrt{\frac{2 GM }{ R }}$
$=\sqrt{\frac{\left[ M ^{-1} L^3 T^{-2}\right][ M ]}{[ L ]}}$
$=\sqrt{\left[ M ^0 L^2 T^{-2}\right]}=\left[ M ^0 L^{ l } T ^{-1}\right]$
$\therefore$ Dimension of left side $=$ Dimension of right side
So, the given formula is correct.
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