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:

We recall the formula for the time period of a satellite in a circular orbit around the Earth. From Kepler’s third law, the period $$T$$ is related to the orbital radius $$r$$ and the mass of the Earth $$M$$ by

$$T \;=\; 2\pi \sqrt{\dfrac{r^{3}}{G\,M}}$$

Here $$G$$ is the universal gravitational constant. We are asked to work with small uncertainties, so we use the rule that if a quantity $$y$$ depends on another quantity $$x$$ as $$y = k\,x^{n}$$, then the relative uncertainty is

$$\dfrac{\Delta y}{y} \;=\; |n| \,\dfrac{\Delta x}{x}.$$

In our case, rewrite the period in the form of a power of $$M$$ alone, because the relative uncertainty in $$r$$ is said to be negligible. We have

$$T = 2\pi\,G^{-1/2}\,r^{3/2}\,M^{-1/2}.$$

The only variable carrying significant uncertainty is the mass $$M$$, and the exponent of $$M$$ in the above expression is $$-\,\dfrac{1}{2}$$. Therefore, the magnitude of the relative uncertainties is connected by

$$\left|\dfrac{\Delta T}{T}\right| \;=\; \dfrac{1}{2}\,\left|\dfrac{\Delta M}{M}\right|.$$

The problem tells us that the relative uncertainty in the period is

$$\dfrac{\Delta T}{T} = 10^{-2}.$$

Substituting this value in the relation gives

$$10^{-2} \;=\; \dfrac{1}{2}\,\dfrac{\Delta M}{M}.$$

Multiplying both sides by $$2$$, we obtain

$$\dfrac{\Delta M}{M} \;=\; 2 \times 10^{-2}.$$

So, the relative uncertainty in the mass of the Earth is $$2 \times 10^{-2}$$.

Hence, the correct answer is Option A.