JKSSB Physics MCQs

Explanation

1. Find the Orbital Radius ($r$)

The acceleration due to gravity at the Earth's surface ($g$) and at the satellite's height ($g'$) are given by:

$$g = \frac{GM}{R^2} = 9.8 \text{ m/s}^2$$

$$g' = \frac{GM}{r^2} = 4.9 \text{ m/s}^2$$

By taking the ratio of these two equations:

$$\frac{g'}{g} = \frac{4.9}{9.8} = \frac{1}{2}$$

$$\frac{GM/r^2}{GM/R^2} = \frac{R^2}{r^2} = \frac{1}{2}$$

$$r^2 = 2R^2 \implies r = \sqrt{2}R$$

Using the given value $\sqrt{2} = 1.414$ and the radius of the Earth $R = 6,371 \text{ km}$:

$$r = 1.414 \times 6371 \text{ km} \approx 9008.594 \text{ km} = 9.0086 \times 10^6 \text{ m}$$

2. Calculate the Orbital Speed ($v$)

For a satellite in a circular orbit, the gravitational force provides the necessary centripetal force:

$$\frac{mv^2}{r} = mg'$$

$$v^2 = g' \times r$$

$$v = \sqrt{g' \times r}$$

Substitute the values $g' = 4.9 \text{ m/s}^2$ and $r = 9.0086 \times 10^6 \text{ m}$:

$$v = \sqrt{4.9 \times 9.0086 \times 10^6}$$

$$v = \sqrt{44.142 \times 10^6}$$

$$v \approx 6.644 \times 10^3 \text{ m/s}$$

$$v \approx 6.6 \text{ km/s}$$

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