Question

The distance of a geo-stationary satellite from the center of the earth (Radius \(\mathrm{R}=6400 \mathrm{~km}\) ) is nearest to (A) \(5 \mathrm{R}\) (B) \(7 \mathrm{R}\) (C) \(10 \mathrm{R}\) (D) \(18 \mathrm{R}\)

Short answer

The distance of a geo-stationary satellite from the center of the Earth is approximately \(6.6 R_E\), where \(R_E\) represents Earth's radius (\(6400\,km\)). Comparing this result with the given options, the nearest option is (B) \(7 R\).

Step-by-step solution

Find the gravitational force acting on the satellite

The gravitational force acting on a satellite can be represented by the formula: \[F = G\frac{m_s m_E}{R^2}\] Where \(F\) is the gravitational force, \(G\) is the gravitational constant \((6.67\times10^{-11}N\cdot m^2/kg^2)\), \(m_s\) is the mass of the satellite, \(m_E\) is the mass of the Earth \((5.97\times10^{24}kg)\), and \(R\) is the distance between the centers of the Earth and the satellite.

Find the centripetal force on the satellite

The centripetal force keeping the satellite in orbit can be represented by the formula: \[F_c = m_s \frac{v^2}{R}\] Where \(F_c\) is the centripetal force, \(m_s\) is the mass of the satellite, \(v\) is the satellite's orbital velocity, and \(R\) is the distance between the centers of the Earth and the satellite. Since the satellite's orbital period is equal to the Earth's rotational period (24 hours), we can find its orbital speed using the formula: \[v = \frac{2\pi R}{T}\] Here, \(T\) is the orbital period which is equal to \(24 \times 3600\,s\).

Equate gravitational force and centripetal force

We can set the gravitational force equal to the centripetal force and solve for R. \[\frac{Gm_s m_E}{R^2} = m_s \frac{v^2}{R}\] Since the mass of the satellite, \(m_s\), appears on both sides of the equation, we can cancel it out: \[\frac{G m_E}{R^2} = \frac{v^2}{R}\]

Solve for R

Substitute the formula for orbital velocity, \(v\), in the equation and solve for R. \[\frac{G m_E}{R^2} = \frac{(2\pi R/T)^2}{R}\] \[\frac{G m_E}{R^2} = \frac{4\pi^2 R^2}{T^2}\] Now, multiply both sides by R^3 and rearrange the equation: \[R^3 = \frac{G m_E T^2}{4\pi^2}\] Plug in the values of G, m_E, and T, and solve for R: \[R^3 = \frac{(6.67\times10^{-11})(5.97\times10^{24})(24\times3600)^2}{4\pi^2}\] \[R \approx 4.22\times10^7 m\] Now, we need to compare this with Earth's radius (\(R_E = 6400\,km = 6.4\times10^6 m\)): \[\frac{R}{R_E} \approx \frac{4.22\times10^7}{6.4\times10^6}\approx 6.6\]

Choose the appropriate option

Based on our calculation, the distance of a geo-stationary satellite from the center of the Earth is closest to \(6.6 R_E\). Comparing this result with the given options, the nearest option is (B) \(7 R\).

Key concepts

Gravitational Force

The gravitational force is a fundamental interaction that ensures celestial bodies in the universe attract each other with a force proportional to their masses. It's governed by Newton's Law of Universal Gravitation, which posits that every object with mass exerts a gravitational pull on every other mass. The force of gravity acting on a satellite orbiting Earth can be expressed as:

• \( F = G\frac{m_s m_E}{R^2} \)

Here, \( G \) is the gravitational constant \( (6.67\times10^{-11}\,N\cdot m^2/kg^2) \), \( m_s \) is the mass of the satellite, \( m_E \) is the mass of Earth \( (5.97\times10^{24}\,kg) \), and \( R \) is the distance between the center of Earth and the satellite. This force diminishes as the square of the distance between the two bodies increases.
Gravitational force is the force that the Earth exerts on any mass that lays on or above its surface. It's what keeps satellites in their respective orbits by pulling them towards Earth's center.

Centripetal Force

Centripetal force is the necessary force that keeps a body moving in a curved path, striving towards a center of rotation. For a satellite, this force is directed towards the center of Earth, maintaining its circular orbit around the planet.

• For satellites, it can be described by:\[ F_c = m_s \frac{v^2}{R} \]
• Where \( F_c \) is the centripetal force, \( m_s \) is the mass of the satellite, \( v \) is the satellite's orbital velocity, and \( R \) is the radius of the orbit.

When a satellite is in a stable orbit, the gravitational force and the centripetal force equalize each other out. This equilibrium means that the satellite doesn't spiral towards the Earth or drift away into space. By maintaining this balance, the satellite remains in its designated geostationary orbit.

Orbital Velocity

Orbital velocity is a critical concept for understanding satellite motion. It is the speed at which a satellite must travel to maintain a stable orbit around Earth without the addition of external thrust.

• This velocity is influenced by the radius of the orbit and the mass of the Earth. It can be calculated by the formula:\[ v = \frac{2\pi R}{T} \]
• In this formula, \( v \) is the orbital velocity, \( R \) is the radius from the center of Earth to the satellite, and \( T \) is the period of orbit (for geostationary satellites, this period equals Earth's rotational period of 24 hours).

Achieving the correct orbital velocity allows a satellite to orbit Earth at the same rate that the Earth is rotating. This synchronization results in the satellite staying above the same geographical point, a necessary condition for a geostationary orbit.

Earth's Radius

Earth's radius is a fundamental unit for calculating the distance of a satellite's orbit, especially for geostationary satellites. The average radius of the Earth is approximately 6400 km or \( 6.4\times10^6 \) meters.
When determining the orbit of geostationary satellites, scientists often compare the satellite's distance from the Earth's center to the Earth's own radius.
This comparison helps in understanding orbital distances in relative terms, aiding engineers and scientists in approximating where these satellites will be positioned relative to Earth.

• For example, as calculated in the given exercise, a geostationary satellite's orbit is about \( 6.6 \) times Earth's radius.

These conversions are valuable when considering options for satellite placement, thus ensuring that these satellites maintain their fixed positions relative to the Earth's surface.