Step 1: Formula for orbital velocity.The velocity (\(v\)) of a satellite in a circular orbit (radius \(r\)), around a planet (mass \(M\)), is found by equating gravitational and centripetal forces:\[ \frac{GMm}{r^2} = \frac{mv^2}{r} $\Rightarrow$ v = \sqrt{\frac{GM}{r}} \]Orbital velocity is inversely proportional to the square root of the orbital radius: \( v \propto \frac{1}{\sqrt{r}} \).
Step 2: Ratio for satellites A and B.\[ \frac{v_B}{v_A} = \frac{\sqrt{GM/r_B}}{\sqrt{GM/r_A}} = \sqrt{\frac{r_A}{r_B}} \]
Step 3: Substitute values and solve for \(v_B\).- \( v_A = 3v \)- \( r_A = 4R \)- \( r_B = R \)\[ \frac{v_B}{3v} = \sqrt{\frac{4R}{R}} = \sqrt{4} = 2 \]\[ v_B = 2 \times (3v) = 6v \]
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)