Question

Two satellites A and B are orbiting a planet in circular orbits with radii 2R and R respectively. If the speed of satellite A is 2v, then the speed of satellite B is

• A. 6√2v
• B. 2√2v
• C. 5√2v
• D. 6v
• E. 4v

Answer 1

The Correct Option is B

The speed \(v\) of a satellite in a circular orbit is given by the formula: \[ v = \sqrt{\frac{GM}{r}} \] Where: - \(G\) is the gravitational constant, - \(M\) is the mass of the planet, - \(r\) is the radius of the orbit. Let the speed of satellite A be \(v_A\) and the speed of satellite B be \(v_B\). For satellite A, the radius of orbit is \(2R\), and for satellite B, the radius is \(R\). From the formula, we know that the speed of satellite A is: \[ v_A = \sqrt{\frac{GM}{2R}} = 2v \] Now, using the same formula for satellite B: \[ v_B = \sqrt{\frac{GM}{R}} \] Using the ratio of the speeds: \[ \frac{v_B}{v_A} = \frac{\sqrt{\frac{GM}{R}}}{\sqrt{\frac{GM}{2R}}} = \sqrt{2} \] Therefore: \[ v_B = v_A \cdot \sqrt{2} = 2v \cdot \sqrt{2} = 2\sqrt{2}v \]
Thus, the speed of satellite B is \( 2\sqrt{2}v \).

The correct option is (B) : \(2\sqrt2v\)

Answer 2

Let the gravitational force provide the necessary centripetal force for a satellite in circular orbit.  

For a satellite of mass m orbiting at radius r around a planet of mass M, the balance of forces gives: 

$$ \frac{G M m}{r^2} = \frac{m v^2}{r} $$ 
Canceling m and simplifying: 
$$ \frac{G M}{r} = v^2 $$ 
So, $$ v = \sqrt{\frac{G M}{r}} $$ 
Speed is inversely proportional to the square root of the radius: $$ v \propto \frac{1}{\sqrt{r}} $$ 
Let the radius of satellite A be \( r_A = 2R \) and speed be \( v_A = 2v \). Let the radius of satellite B be \( r_B = R \) and speed be \( v_B \). 

Using the proportionality: $$ \frac{v_B}{v_A} = \sqrt{\frac{r_A}{r_B}} = \sqrt{\frac{2R}{R}} = \sqrt{2} $$ 
So, $$ v_B = v_A \cdot \sqrt{2} = 2v \cdot \sqrt{2} = 2\sqrt{2}v $$ 
Correct answer: 2√2v