If two satellites of masses m1 and m2 are revolving around a earth in a circular orbits of radius r1 and r2, then the ratio of their orbital velocities \(\dfrac{v_1}{v_2}\) is
When satellites revolve around the Earth in circular orbits, their speed, known as orbital velocity, depends on the mass of the Earth and the radius of their orbit. The masses of the satellites themselves do not directly affect their orbital velocity in a given orbit.
The orbital velocity \( v \) of a satellite in a circular orbit around a planet (like Earth) is given by the formula:
\( v = \sqrt{\dfrac{GM}{r}} \)
Where:
Notice that the mass of the satellite itself (\( m_1 \) or \( m_2 \)) does not appear in this formula. This means the orbital velocity depends only on the mass of the planet and the distance from the center of the planet to the satellite.
We have two satellites, satellite 1 with mass \( m_1 \) in an orbit of radius \( r_1 \), and satellite 2 with mass \( m_2 \) in an orbit of radius \( r_2 \). Using the orbital velocity formula:
The orbital velocity of satellite 1 is \( v_1 = \sqrt{\dfrac{GM}{r_1}} \).
The orbital velocity of satellite 2 is \( v_2 = \sqrt{\dfrac{GM}{r_2}} \).
To find the ratio of their orbital velocities, \( \dfrac{v_1}{v_2} \), we divide the expression for \( v_1 \) by the expression for \( v_2 \):
\( \dfrac{v_1}{v_2} = \dfrac{\sqrt{\dfrac{GM}{r_1}}}{\sqrt{\dfrac{GM}{r_2}}} \)
We can combine the square roots:
\( \dfrac{v_1}{v_2} = \sqrt{\dfrac{\dfrac{GM}{r_1}}{\dfrac{GM}{r_2}}} \)
Now, simplify the fraction inside the square root. Dividing by a fraction is the same as multiplying by its reciprocal:
\( \dfrac{v_1}{v_2} = \sqrt{\dfrac{GM}{r_1} \times \dfrac{r_2}{GM}} \)
The \( GM \) terms cancel out:
\( \dfrac{v_1}{v_2} = \sqrt{\dfrac{\cancel{GM}}{r_1} \times \dfrac{r_2}{\cancel{GM}}} \)
\( \dfrac{v_1}{v_2} = \sqrt{\dfrac{r_2}{r_1}} \)
Therefore, the ratio of their orbital velocities \( \dfrac{v_1}{v_2} \) is equal to the square root of the inverse ratio of their orbital radii, \( \sqrt{\dfrac{r_2}{r_1}} \). This calculation confirms how the orbital velocity ratio depends only on the radii.
This step-by-step process shows how the ratio of orbital velocities is derived directly from the fundamental formula.
Which one of the following statement is true for the relation, \(F= \frac{{G{m_1}{m_2}}}{{{r^2}}}\) ?
(All symbols have their usual meanings)The free-fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is
A planet has a mass M 1and radius R 1. The value of acceleration due to gravity on its surface is g 1. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?
Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be
Suppose the force of gravitation between two bodies of equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become