Problem 771 - JEE Physics

Question

A satellite is moving around the earth with speed \(\mathrm{v}\) in a circular orbit of radius \(\mathrm{r}\). If the orbit radius is decreased by \(1 \%\) its speed will (A) increase by \(1 \%\) (B) increase by \(0.5 \%\) (C) decreased by \(1 \%\) (C) Decreased by \(0.5 \%\)

Step-by-Step Solution

Answer

The speed of the satellite will increase by approximately \(0.5\%\) when the orbit radius is decreased by \(1\%\).

Step 1: Write down the formula for gravitational force and centripetal force

The formula for gravitational force acting on an object is given by \(F_g = G \frac{Mm}{r^2}\), where G is the universal gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between their centers. The formula for centripetal force acting on an object moving in a circle of radius r and speed v is given by \(F_c = mv^2/r\), where m is the mass of the satellite and v is the speed of the satellite.

Step 2: Equate gravitational force and centripetal force

Since the satellite is moving in a circular orbit, the gravitational force acting on it will be equal to the centripetal force. Therefore, we can equate the two formulas: \(G \frac{Mm}{r^2} = m \frac{v^2}{r}\)

Step 3: Solve for v in terms of r

We can cancel \(\frac{1}{r}\) on both sides and divide by m to find an expression for v^2: \(\frac{GM}{r} = v^2\) Take square root on both sides: \(v = \sqrt{\frac{GM}{r}}\)

Step 4: Calculate the new values for r and v

When the radius of the orbit is decreased by 1%, the new radius will be 0.99r. We can use the derived expression for v to find the new speed of the satellite: \(v' = \sqrt{\frac{GM}{0.99r}}\)

Step 5: Calculate the percentage change in speed

We can now find the percentage change in speed as follows: \(\% \Delta v = \frac{v' - v}{v} \times 100\) Substitute the expressions for v and v': \(\% \Delta v = \frac{\sqrt{\frac{GM}{0.99r}} - \sqrt{\frac{GM}{r}}}{\sqrt{\frac{GM}{r}}} \times 100\) Simplify the expression: \(\% \Delta v = \left(\frac{\sqrt{\frac{1}{0.99}} - 1}{1}\right) \times 100\)

Step 6: Calculate the numerical value

Using a calculator, we find the percentage change in speed to be approximately: \(\% \Delta v \approx 0.503 \%\) So, the answer is (B) increase by 0.5%.

Key Concepts

Gravitational Force

Gravitational force is a fundamental force of nature that acts between two objects with mass. It is responsible for keeping planets in orbit around stars, like Earth around the Sun, and satellites around planets.

When discussing a satellite orbiting Earth, the gravitational force is the attractive pull that Earth exerts on the satellite. The strength of this force can be calculated using the formula:

\( F_g = G \frac{Mm}{r^2} \)

• \(G\) is the universal gravitational constant.
• \(M\) is the mass of Earth.
• \(m\) is the mass of the satellite.
• \(r\) is the distance between the center of Earth and the satellite.

Once this force is known, it can be used to predict and understand the behavior of satellites in various orbits.

Centripetal Force

Centripetal force is the necessary force that keeps an object moving in a circular path. It acts perpendicular to the motion of the object and towards the center of the circle.

For a satellite in a circular orbit, the centripetal force is provided by the gravitational pull of Earth. It can be expressed as:

\( F_c = \frac{mv^2}{r} \)

• \(m\) is the mass of the satellite.
• \(v\) is the speed of the satellite.
• \(r\) is the radius of the circular orbit.

Without this force, a satellite would not be able to maintain its circular path and would fly off into space due to inertia.

Orbital Mechanics

Orbital mechanics is the study of the motions of artificial satellites and natural celestial bodies. It involves understanding how forces like gravity and inertia interact to create the various types of orbits we observe.

In the context of a satellite, orbital mechanics allows us to calculate important parameters such as orbit speed, period, and radius. By equating the gravitational and centripetal forces, as shown in the solution steps, we derive a critical relationship for circular motion:

\( \frac{GM}{r} = v^2 \)

This equation indicates that the velocity of a satellite in orbit is dependent on the gravitational pull of Earth and the radius of the orbit. Understanding these principles is vital for tasks such as launching satellites and planning space missions.

Circular Orbit

A circular orbit is one where the satellite travels in a perfect circle around the Earth, maintaining a constant altitude and speed. This is a special case in orbital mechanics and is easier to analyze compared to elliptical or other orbit shapes.

In a circular orbit, the speed of the satellite can be found using the formula:

\( v = \sqrt{\frac{GM}{r}} \)

Here, the speed remains constant as long as the radius is unchanged. However, a slight change in the orbit radius, as the exercise involves reducing by 1%, affects the speed.

• As the radius decreases, the speed increases to maintain the balance of forces.
• This relationship ensures that the gravitational and centripetal forces remain equal.

Analyzing these changes provides insights into how satellites adjust to different orbits, crucial for satellite operations and missions.