Question

Two satellites of mass \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\left(\mathrm{~m}_{1}=\mathrm{m}_{2}\right)\) are revolving round the earth in circular orbits of \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\left(\mathrm{r}_{1}>\mathrm{r}_{2}\right)\) respectively. Which of the following statement is true regarding their speeds \(\mathrm{V}_{1}\) and \(\mathrm{V}_{2}\) (A) \(\mathrm{V}_{1}=\mathrm{V}_{2}\) (B) \(\mathrm{V}_{1}<\mathrm{V}_{2}\) (C) \(\mathrm{V}_{1}>\mathrm{V}_{2}\) (D) \(\left(\mathrm{V}_{1} / \mathrm{r}_{1}\right)=\left(\mathrm{V}_{2} / \mathrm{r}_{2}\right)\)

Short answer

The correct statement regarding the velocities of the two satellites is (B) \(V_1 < V_2\).

Step-by-step solution

Recall the gravitational force formula

The gravitational force that acts on the two satellites can be expressed using the formula: \[F = \dfrac{G \cdot m_1 \cdot m_2}{r^2}\] Where: - \(F\) is the gravitational force acting on the satellite - \(G\) is the gravitational constant - \(m_1\) and \(m_2\) are the masses of the satellite and Earth respectively - \(r\) is the distance between the centers of mass of the satellite and the Earth

Recall the formula for centripetal force

The centripetal force required to keep the satellite in a circular orbit around the Earth is given by: \[F_c = m\cdot V^2 / r\] Where: - \(F_c\) is the centripetal force acting on the satellite - \(m\) is the mass of the satellite - \(V\) is the velocity of the satellite - \(r\) is the radius of the circular orbit

Equate gravitational force and centripetal force for the satellites

Since the gravitational force is what provides the centripetal force needed to keep the satellite in its circular orbit, we can equate the two formulas: \[\dfrac{G \cdot m_1 \cdot m_2}{r^2} = m\cdot V^2 / r\] Now we need to compare the two situations with satellites \(1\) and \(2\): 1. For satellite \(1\): \[\dfrac{G \cdot m_1 \cdot m_E}{r_1^2} = m_1 \cdot V_1^2 / r_1\] 2. For satellite \(2\): \[\dfrac{G \cdot m_1 \cdot m_E}{r_2^2} = m_1 \cdot V_2^2 / r_2\]

Divide the equations for the two satellites to compare their velocities

Divide equation \(1\) by equation \(2\): \[\dfrac{\frac{G \cdot m_1 \cdot m_E}{r_1^2}}{\frac{G \cdot m_1 \cdot m_E}{r_2^2}} = \dfrac{\frac{m_1 \cdot V_1^2}{r_1}}{\frac{m_1 \cdot V_2^2}{r_2}}\] Simplify by canceling out \(G\), \(m_1\), and \(m_E\): \[\dfrac{r_2^2}{r_1^2} = \dfrac{V_1^2}{V_2^2} \cdot \dfrac{r_2}{r_1}\]

Find the relationship between the velocities of the two satellites

Rearrange the equation to find the relationship between \(V_1\) and \(V_2\): \[\dfrac{V_1^2}{V_2^2} = \dfrac{r_1}{r_2}\] Since \(r_1 > r_2\), it follows that: \(V_1^2 < V_2^2\) Taking the square root of both sides: \(V_1 < V_2\) Conclusion: The correct statement regarding the velocities of the two satellites is: (B) \(V_1 < V_2\)

Key concepts

Satellites

Satellites are objects that orbit around a larger body in space due to the gravitational pull of the larger body. In this exercise, we consider two satellites orbiting the Earth in circular paths. These satellites remain in orbit because of the perfect balance between their forward motion and the gravitational pull of the Earth.

Here are key points to understand about satellites:

• They can be natural, like moons, or artificial, like the International Space Station.
• The speed a satellite travels depends on its distance from Earth. This is because the gravitational force changes with distance.
• Satellites at different heights from Earth have different orbital velocities.

Satellites revolve around Earth with velocities calculated based on their orbital radius and other factors affecting gravitational force. The higher a satellite, the slower it travels in its orbit compared to a satellite closer to Earth. Understanding these dynamics helps in planning satellite missions and maintaining their paths.

Centripetal Force

Centripetal force is the inward force required to keep an object moving in a circular path. For satellites, this force is necessary to keep them in orbit around Earth. The centripetal force is provided by Earth's gravitational force.

Consider these points when thinking about centripetal force:

• This force is always directed toward the center of the circular path.
• It prevents the revolving object, like a satellite, from flying off into space.
• The amount of centripetal force needed depends on the mass of the object, its velocity, and the radius of its path.

The centripetal force for a satellite, represented by the equation \[F_c = m \cdot V^2 / r\],demonstrates that increasing the velocity or decreasing the orbit's radius increases the centripetal force needed. Satellites orbiting close to Earth require greater velocity to maintain their trajectory as they need to overcome stronger gravitational forces.

Gravitational Force

Gravitational force is the attraction between two masses. In the context of satellites, it is the force that keeps them in their orbit around Earth. Since gravitational force acts along the line joining the centers of the two masses, it provides the necessary pull towards Earth, maintaining the satellite on its curved path.

Key ideas about gravitational force:

• It is proportional to the product of the two masses involved.
• It diminishes with the square of the distance between the two masses.
• It is universal, affecting all objects with mass, irrespective of their size.

The gravitational force equation is:\[F = \dfrac{G \cdot m_1 \cdot m_2}{r^2}\],where \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the two bodies (e.g., the satellite and the Earth), and \(r\) is the distance between their centers. When this force equals the centripetal force needed, the satellite maintains its stable orbit. This delicate balance ensures satellites neither drift into space nor crash back to Earth, allowing them to effectively serve functions like communication, navigation, and observation.