Problem 786 - JEE Physics

Question

Two satellites \(\mathrm{A}\) and \(\mathrm{B}\) go round a planet in circular orbits having radii \(4 \mathrm{R}\) and \(\mathrm{R}\) respectively If the speed of satellite \(\mathrm{A}\) is \(3 \mathrm{v}\), then speed of satellite \(\mathrm{B}\) is (A) \((3 \mathrm{v} / 2)\) (B) \((4 \mathrm{v} / 2)\) (C) \(6 \mathrm{v}\) (D) \(12 \mathrm{v}\)

Step-by-Step Solution

Answer

The speed of satellite B is \(6v\). The answer is (C) \(6 \mathrm{v}\).

Step 1: Gravitational force

In order to maintain a satellite's orbit around a planet, the gravitational force acting on the satellite must be equal to the centripetal force required for the satellite's circular motion. This can be expressed by the following equation: \[F_g = F_c\] The gravitational force is given by Newton's law of universal gravitation: \[F_g = G \frac{m_pm_s}{r^2}\] where: - \(F_g\) is the gravitational force, - \(G\) is the gravitational constant, - \(m_p\) is the mass of the planet, - \(m_s\) is the mass of the satellite, and - \(r\) is the distance between the satellite and the center of the planet (i.e., the radius of the satellite's orbit).

Step 2: Centripetal force and acceleration

The centripetal force is given by the following formula: \[F_c = m_s a_c\] where: - \(F_c\) is the centripetal force, and - \(a_c\) is the centripetal acceleration. The centripetal acceleration can be expressed in terms of the satellite's speed as follows: \[a_c = \frac{v^2}{r}\]

Step 3: Equating gravitational force and centripetal force

Now let's equate the expressions for gravitational force and centripetal force: \[G \frac{m_pm_s}{r^2} = m_s \frac{v^2}{r}\] We can cancel out \(m_s\) from both sides: \[G \frac{m_p}{r^2} = \frac{v^2}{r}\] Now, let's solve for the speed of each satellite in terms of the radius of their orbits and the mass of the planet: \[v^2 = G \frac{m_p r}{r^2}\] \[v = \sqrt{G \frac{m_p}{r}}\]

Step 4: Finding the speed ratio of satellite A and satellite B

Since we are given the speed of satellite A, we can find the speed of satellite B by finding the ratio of their speeds. Let A's speed be \(3v\), and the ratio of A's radius to B's radius is given as \(4R : R\). Now we can write the speed of satellite A and satellite B as follows: \[v_A = 3v = \sqrt{G \frac{m_p}{4R}}\] \[v_B = \sqrt{G \frac{m_p}{R}}\] Taking the ratio of their speeds: \[\frac{v_B}{v_A} = \frac{\sqrt{G \frac{m_p}{R}}}{\sqrt{G \frac{m_p}{4R}}}\] We can simplify this expression: \[\frac{v_B}{v_A} = \frac{\sqrt{4}}{\sqrt{1}} = 2\] Now, we can find the speed of satellite B: \[v_B = 2 \times v_A = 2 \times 3v = 6v\] Thus, the speed of satellite B is \(6v\). The answer is (C) \(6 \mathrm{v}\).

Key Concepts

Gravitational Force

Gravitational force is a fundamental concept in orbital mechanics, describing how two masses attract each other. This force keeps satellites in orbit around planets. According to Newton's law of universal gravitation, the gravitational force (\(F_g\)) between two objects is:

• Directly proportional to the product of their masses
• Inversely proportional to the square of the distance between their centers

The formula is:\[F_g = G \frac{m_p m_s}{r^2}\]where \(G\) is the gravitational constant, and \(m_p\) and \(m_s\) are the masses of the planet and satellite, respectively. \(r\) stands for the orbital radius. This attractive force ensures that satellites remain in stable orbits when balanced with the centripetal force needed to navigate a circular path.

Centripetal Force

For a satellite to maintain its circular path, it requires a centripetal force to act inward toward the center of its path. Imagine swinging a ball attached to a string in a circle; the tension in the string acts as the centripetal force. Similarly, in satellite motion, gravitational force acts as the centripetal force.The centripetal force (\(F_c\)) needed for circular motion is calculated using:\[F_c = m_s a_c\]where \(a_c\) is the centripetal acceleration, and it can be represented as:\[a_c = \frac{v^2}{r}\]Here, \(v\) is the speed of the satellite, and \(r\) is the radius of the circular orbit. For stable orbits, the forces are balanced as \(F_g = F_c\), ensuring the satellite follows its path without spiraling in or out.

Satellite Motion

Satellite motion refers to the way satellites orbit around a celestial body, such as Earth, in a stable path. The balance of gravitational and centripetal forces dictates this motion. Each satellite's speed and orbital radius determine how these forces interact. Unlike linear motion, satellites move in elliptical or circular paths due to the gravitational pull. Satellites are influenced by:

• Gravitational attraction between the satellite and the planet
• Orbital speed, which must be sufficient to counteract gravity
• Altitude or orbit radius, affecting potential energy and speed

When calculating satellite motion, focus on the orbital velocity formula. For different orbits, the gravitational pull and necessary speed must be calculated to ensure satellites are kept in a stable path in harmony with gravitational and centripetal forces.

Circular Orbits

Circular orbits are specific paths where satellites travel at constant speed around a planet at a fixed distance, forming a circle. These orbits are easy to understand and calculate, serving as a useful starting point in orbital mechanics.For a satellite in a circular orbit, the balance of \(F_g\) and \(F_c\) ensures a consistent radius and velocity:\[v = \sqrt{G \frac{m_p}{r}}\]In our exercise, satellites A and B orbit with different radii, \(4R\) and \(R\) respectively. This impacts their orbital speed. Satellite B, closer to the planet, must move faster to maintain its path given its smaller orbit. Understanding circular orbits helps determine factors like speed, stability, and period, making it crucial for efficient satellite deployment and navigation.