JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 17)
The relative uncertainly in the period of a satellite orbiting around the earth is 10-2. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :
10$$-$$2
2 $$ \times $$ 10$$-$$2
3 $$ \times $$ 10$$-$$2
6 $$ \times $$ 10$$-$$2
Explanation
From kepler's law,
T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$
$$ \Rightarrow $$$$\,\,\,$$T2 = $${{4{\pi ^2}} \over {GM}}{r^3}$$
$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$
$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\Delta T} \over T}$$ + 3 $${{\Delta r} \over r}$$
as $${{\Delta r} \over r} \simeq 0$$
$$\therefore\,\,\,$$ $$\left| {{{\Delta M} \over M}} \right|$$ = 2 $${{\Delta T} \over T}$$ = 2 $$ \times $$ 10$$-$$2
T = 2$$\pi $$ $$\sqrt {{{{r^3}} \over {GM}}} $$
$$ \Rightarrow $$$$\,\,\,$$T2 = $${{4{\pi ^2}} \over {GM}}{r^3}$$
$$ \Rightarrow $$$$\,\,\,$$ M = $${{4{\pi ^2}} \over G} \times {{{r^3}} \over {{T^2}}}$$
$$\therefore\,\,\,$$ $${{\Delta M} \over M}$$ = 2 $${{\Delta T} \over T}$$ + 3 $${{\Delta r} \over r}$$
as $${{\Delta r} \over r} \simeq 0$$
$$\therefore\,\,\,$$ $$\left| {{{\Delta M} \over M}} \right|$$ = 2 $${{\Delta T} \over T}$$ = 2 $$ \times $$ 10$$-$$2
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