Orbits of Satellites and Planets

Orbits

When looking at the orbits of planets and satellites, it is the gravitational force that acts as the centripetal force keeping the planet or satellite at a fixed radius of orbit.

Time Period of Orbit

We know \: \text{velocity} = \dfrac{\text{distance}}{\text{time}}. The circumference of a satellite in circular motion is equal to 2 \pi r where r is the radius of orbit. So we can say that the velocity of a satellite in circular orbit is:

v = \dfrac{2 \pi r}{T}

where T is the time it takes the satellite to orbit once (the period). We can substitute this into our previous equation for the velocity needed by a satellite to maintain a constant orbit:

\left(\dfrac{2 \pi r}{T}\right)^2 = \dfrac{GM}{r}

This rearranged becomes:

T^2 = \dfrac{4 \pi^2 r^3}{GM}

Kepler’s third law can be derived from this equation which states that the square of the time period of a planet in circular motion is proportional to the cube of the radius:

T^2 \propto r^3

Example: Calculate the time period of the satellite that orbits the Earth at a height of 2500 \: \text{m} above the Earth’s surface.

[3 marks]

r = 6.37 \times 10^6 + 2500 = 6372500 \: \text{m}

T^2 = \dfrac{4 \pi^2 r^3}{GM}

T = \sqrt{\dfrac{4 \pi ^2 \times 6372500^3}{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}}

T = 5070 \: \text{s} \left(\text{3 s.f}\right)

Energy of an Orbiting Satellite

 

Escape Velocity

Escape velocity can be defined as the velocity needed by an object to escape the gravitational field that the object is within.

The escape velocity is independent of the mass of the object but dependent upon the mass of the planet and distance away from the centre of the Earth.

It is important to note that the escape velocity calculation is for objects that do not receive an energy input once leaving the surface of the planet. Therefore, a conventional rocket which has boosters, would not need to be calculated in this way. An object’s escape velocity can be calculated by equating its kinetic energy to gravitational potential energy:

\text{Kinetic Energy} = \text{Gravitational Potential Energy}

\dfrac{1}{2} m v^2 = \dfrac{GMm}{r}

Rearranging these equations and cancelling out the mass of the object \left(m\right) gives:

v = \sqrt{\dfrac{2GM}{r}}

Example: An object is leaving the Earth’s surface. Given that the radius of the Earth is 6.37 \times 10^{6} \: \text{m} and mass 5.97 \times 10^{24} \: \text{kg} and the gravitational constant \left(G\right) is 6.67 \times 10^{-11}, calculate the escape velocity needed for the object.

[2 marks]

v = \sqrt{\dfrac{2GM}{r}}

v = \sqrt{\dfrac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^6}}

v = 11.2 \times 10^3 \: \text{ms}^{-1}

Types of Orbit

Different satellites have different types of orbit depending upon their purpose.

Synchronous Orbits:

If an orbit is synchronous, the orbiting object such as a satellite orbits with the same time period and in the same direction as the planet it is orbiting. To achieve this, the velocity of the satellite must be greater than the velocity of the planet and the greater the radius of the satellite, the greater the velocity would need to be.