Question

A remote sensing satellite of earth revolves in a circular orbit at a height of 0.25 × $10^6$ m above thesurface of earth. Ifearth's radius is 6.38 × $10^6$ m, then the orbital speed of the satellite is

• A. 8.56 km $s^{-1}$
• B. 9.13 km $s^{-1}$
• C. 6.67 km $s^{-1}$
• D. 7.76 km $s^{-1}$

Answer 1

Answer: (D)

7.76 km $s^{-1}$

Answer 2

The orbital speed $v$ of a satellite in a circular orbit is given by:

$$v = \sqrt{\frac{GM}{r}}$$

where:

$G$ is the gravitational constant,

$M$ is the mass of the Earth (with $GM \approx 3.986 \times 10^{14} \, \text{m}^3/\text{s}^2$), and

$r$ is the distance from the center of the Earth.

The satellite's altitude is given as $0.25 \times 10^6 \, \text{m}$ and the Earth's radius is $6.38 \times 10^6 \, \text{m}$. Thus,

$$r = 6.38 \times 10^6 + 0.25 \times 10^6 = 6.63 \times 10^6 \, \text{m}.$$

Now, plugging in the values:

$$v = \sqrt{\frac{3.986 \times 10^{14}}{6.63 \times 10^6}}.$$

Calculating the inside term:

$$\frac{3.986 \times 10^{14}}{6.63 \times 10^6} \approx 6.01 \times 10^7 \, \text{m}^2/\text{s}^2.$$

Taking the square root:

$$v \approx \sqrt{6.01 \times 10^7} \approx 7.75 \times 10^3 \, \text{m/s} \approx 7.76 \, \text{km/s}.$$