Two satellites A and B rotate round a planet’s orbit having radius 4R and R respectively. If the speed of satellite A is 3 V then speed of satellite B is
\(\frac {3V}{2}\)
6V
\(\frac {4V}{2}\)
12 V
The speed of an object in a circular orbit is given by the formula:
v = \(\sqrt{\frac {GM}{R}}\)
Let's compare the speeds of satellites A and B: For satellite A:
vA = \(\sqrt{\frac {GM}{4R}}\)
For satellite B:
vB = \(\sqrt{\frac {GM}{R}}\)
vA = \(\sqrt{\frac {GM}{4R}}\)
vA = \(\sqrt{\frac {1}{4}}\) \(\sqrt{\frac {GM}{R}}\)
vA = \(\frac {1}{2}\)vB
Therefore, the speed of satellite B is twice the speed of satellite A, which means:
vB = 2 vA
vB = 2 x 3V
vB = 6V
So, the correct option is (B) 6V.
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
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.)