The relative uncertainty 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 :

Question

The relative uncertainty 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 : (A) 10-2 (B)  2 × 10-2 (C) 3 × 10-2 (D) 6 × 10-2

Tardigrade Admin · Accepted Answer

Time period of revolutions of a satellite revolving around Earth of mass

M

in orbit of radius

r

is given by

T = 2 π \frac{r ^{3}}{GM}

\Rightarrow T^{2} = \frac{4 π ^{2} r ^{3}}{GM}

Taking log of both sides, we get

lo g T^{2} = lo g \frac{4 π ^{2} r ^{3}}{GM} = lo g \frac{4 π ^{2} r ^{3}}{G} - lo g (M)

\Rightarrow 2 lo g T = lo g k - lo g M

\Rightarrow 2 \frac{Δ T}{T} = \frac{Δ M}{M}

\Rightarrow \frac{Δ M}{M} = 2 \times 1 0^{- 2}

(

∵

Given relative uncertainty in period,

\frac{Δ T}{T} = 1 0^{- 2}

)

Q. The relative uncertainty in the period of a satellite orbiting around the earth is $1 0^{- 2}$ . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

$1 0^{- 2}$

$2 \times 1 0^{- 2}$

$3 \times 1 0^{- 2}$

$6 \times 1 0^{- 2}$

Q. The relative uncertainty 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

2×10−2

3×10−2

6×10−2

Q. The relative uncertainty in the period of a satellite orbiting around the earth is $1 0^{- 2}$ . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

$1 0^{- 2}$

$2 \times 1 0^{- 2}$

$3 \times 1 0^{- 2}$

$6 \times 1 0^{- 2}$