# Question 1068c

Jun 5, 2016

$\text{Option D"->} \approx 7.4 \frac{k m}{s}$

#### Explanation:

Let

• $m \to \text{Mass of satellite}$
• $M \to \text{Mass of Earth}$
• $R \to \text{Radius of Earth} = 6400 k m$
• $h \to \text{Height of the orbit from surface of the Earth} = 900 k m$
• $v \to \text{Orbiting speed of the satellite}$
• $G \to \text{Gravitational constant}$
• $g \to \text{Acceleration due to gravity} = 9.8 \times {10}^{-} 3 k m {s}^{-} 2$

Now we know

$\text{Centripetal pull on earth " = "Gravitational force acting between}$

$\implies \frac{m {v}^{2}}{R + h} = \frac{G m M}{R + h} ^ 2$

$\implies \frac{\cancel{m} {v}^{2}}{\cancel{R + h}} = \frac{G \cancel{m} M}{R + h} ^ \cancel{2}$

=>v^2=(GM)/(R+h))

$\implies {v}^{2} = \frac{g {R}^{2}}{R + h} \ldots \ldots \left(1\right)$ " "[ Since " " GM=gR^2]#

Now inserting the values in equation (1) we get

$\implies {v}^{2} = \frac{9.8 \times {10}^{-} 3 {\left(6400\right)}^{2}}{6400 + 900}$

$\implies v = \sqrt{\frac{9.8 \times {10}^{-} 3 {\left(6400\right)}^{2}}{6400 + 900}} \approx 7.4 \frac{k m}{s}$