Two satellites \(A\) and \(B\) go round a planet in circular orbits having radii \(4R\) and \(R\) respectively. If the speed of \(A\) is \(3v\), the speed of \(B\) will be:
To solve this problem, we need to understand the principles governing the motion of satellites in circular orbits. According to Kepler's laws and gravitational principles, the centripetal force required to keep a satellite in orbit is provided by the gravitational force. This sets up a relationship between the speed of the satellite, the radius of its orbit, and the mass of the planet being orbited.
For a satellite in a circular orbit, the gravitational force \(F_g\) provides the necessary centripetal force \(F_c\):
\(F_g = \frac{G M m}{r^2}\)
and
\(F_c = \frac{m v^2}{r}\)
where \(G\) is the gravitational constant, \(M\) is the mass of the planet, \(m\) is the mass of the satellite, \(r\) is the radius of the orbit, and \(v\) is the speed of the satellite.
By equating the gravitational force to the centripetal force, we get:
\(\frac{G M m}{r^2} = \frac{m v^2}{r}\)
Solving for \(v\), the velocity of each satellite, we find:
\(v = \sqrt{\frac{G M}{r}}\)
This equation shows that the speed of a satellite is inversely proportional to the square root of the radius of its orbit, \(r\).
Given:
Radius of orbit for satellite \(A\), \(r_A = 4R\), and its speed \(v_A = 3v\).
The orbital speed of a satellite is given by: \[v = \sqrt{\frac{GM}{R}}.\] For satellites \(A\) and \(B\): \[\frac{v_A}{v_B} = \sqrt{\frac{R_B}{R_A}} = \sqrt{\frac{R}{4R}} = \frac{1}{2}.\] Thus: \[v_B = 2v_A.\] Given \(v_A = 3v\), the speed of \(B\) is: \[v_B = 2 \cdot 3v = 6v.\]
