Problem 768 - JEE Physics

Question

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 8 (B) 4 (C) 6 (D) \(11.2\)

Step-by-Step Solution

Answer

The orbital velocity of a small satellite revolving near the Earth's surface is approximately 8 \(\mathrm{km/s}\).

Step 1: Write the formula for orbital velocity

Orbital velocity (V) is the velocity an object requires to stay in a stable orbit around another celestial body, like a planet or a moon. The formula for orbital velocity is given by: \[V = \sqrt{\frac{GM}{R}}\] Where: - V is the orbital velocity - G is the gravitational constant (\(6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \cdot \mathrm{kg}^{-2}\)) - M is the mass of the celestial body (in this case, Earth's mass, which is \(5.97 \times 10^{24} \mathrm{kg}\)) - R is the distance from the center of the celestial body to the object in orbit (in this case, Earth's radius, which is approximately \(6.37 \times 10^6 \mathrm{m}\))

Step 2: Substitute values and calculate the orbital velocity

We'll now substitute the given values for G, M, and R into the orbital velocity formula: \[V = \sqrt{\frac{(6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \cdot \mathrm{kg}^{-2})(5.97 \times 10^{24} \mathrm{kg})}{(6.37 \times 10^6 \mathrm{m})}}\] Now we'll solve for V: \[V \approx 7.9 \times 10^3 \mathrm{m/s}\]

Step 3: Convert the orbital velocity to \(\mathrm{km/s}\) and select the correct answer

Wrap up the process by converting the orbital velocity from \(\mathrm{m/s}\) to \(\mathrm{km/s}\) as requested: \[V \approx 7.9 \times 10^3 \mathrm{m/s} \times (\frac{1 \mathrm{km}}{1000 \mathrm{m}}) \approx 7.9 \mathrm{km/s}\] Looking at the given choices, the closest option to our calculated orbital velocity of \(7.9 \mathrm{km/s}\) is choice (A) 8 \(\mathrm{km/s}\). So, the correct answer is: (A) 8 \(\mathrm{km/s}\)

Key Concepts

Gravitational Constant

The gravitational constant, often represented by the symbol \( G \), plays a crucial role in calculations involving gravity, especially in celestial mechanics. It is a proportionality factor used in Newton's Law of Universal Gravitation, which describes the attractive force between two masses.

• The value of the gravitational constant is \( 6.67 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^2 \cdot \mathrm{kg}^{-2} \).
• This value is tiny, reflective of the feeble nature of gravitational force compared to other fundamental forces.
• \( G \) is used universally, providing a consistent measure that helps us understand the gravitational effects in cosmic scale calculations.

In our formula for orbital velocity, \( G \) allows us to relate the mass of Earth (or any celestial body) and the distance to the object in orbit to calculate the necessary velocity to maintain a stable orbit.

Mass of Earth

The mass of Earth is a critical parameter when calculating the orbital velocity of a satellite. It represents the total mass of our planet and is a central component in the formula governing gravitational forces between the Earth and any object in its vicinity.

• Earth’s mass is approximately \( 5.97 \times 10^{24} \mathrm{kg} \).
• This huge mass contributes significantly to its gravitational pull, affecting both natural satellites and human-made ones.
• Understanding Earth's mass helps scientists and engineers to determine the capabilities and limits of satellite motion.

In our calculations, Earth’s mass is used alongside the gravitational constant to determine the necessary velocity for a satellite to orbit near Earth's surface without falling back to the ground or drifting into space.

Radius of Earth

The radius of Earth is a vital dimension in the formula for orbital velocity. It is the average distance from Earth's center to its surface and helps determine the gravitational field strength experienced by an object in orbit.

• The average radius of Earth is approximately \( 6.37 \times 10^6 \mathrm{m} \) or \( 6,370 \ \mathrm{km} \).
• This measurement is key to calculating the orbital path and speed necessary to maintain orbit close to Earth's surface.
• When a satellite orbits at this altitude, it benefits from a balance between gravitational pull and the centrifugal force due to its velocity.

Calculating orbital velocity requires knowing both the radius and mass of Earth, together with the gravitational constant, to ensure accurate and safe operation of satellites.

Satellite Motion

Satellite motion refers to how satellites move in space, particularly around larger celestial bodies like Earth. These movements are governed by the principles of orbital mechanics and are influenced by gravitational forces.

• To remain in orbit, a satellite must travel at a precise speed, known as orbital velocity, which ensures the gravitational pull from the Earth is perfectly balanced by the satellite's inertia.
• If the velocity is too low, the satellite will fall back to Earth; if it's too high, it will escape into space.
• An orbital velocity near Earth’s surface is approximately \( 7.9 \ \mathrm{km/s} \), allowing satellites to stay aloft while orbiting in a stable path.

Understanding these dynamics is essential for satellite design and launch, allowing technology like weather satellites, GPS, and communication devices to function reliably and efficiently.