WHAT IS NASA PHYSICS?
MODULES
Forces and Motion
Conservation of Momentum & Energy
Temperature and Heat
Fluids
Optics
Electromagnetic Spectrum
Modern Physics
Anticipation Guide 7
Intro to Modern Physics
Blackbody Radiation
The Ultraviolet Catastrophe
The Photoelectric Effect
Bohr's Atom
Spectra
Radioactive Decay
Special Relativity (SR)
Simultaneity
Distance and Time
General Relativity
May the Forces be with You
Modern Physics Notebook
Assessment Problems 7
Useful Things
SITE MAP
Forces and Motion
Anticipation Guide
Speed and Velocity
Acceleration
Gravity
Projectile Motion
Orbital Motion
Newton's Laws of Motion
Assessment Problems
Equation
v = (GM/r)^{½}
1.16
Reminder
G= 6.673 x 10^{-11} Nm^{2}/kg^{2}
Orbital Motion
Orbital Velocity
As we have seen, an object must reach a critical tangential velocity in order to go into orbit around a planet. An object would continue in a straight line forever unless some force acted on it. This force is gravity. Gravity pulls. It pulls the potentially orbiting object towards the center of the planet.
Taking the example of the Earth and a satellite payload that was just launched, we want to send the payload fast enough tangentially or, in other words, perpendicular to the force of gravity such that it will always miss the Earth’s surface as it falls. To do this, it has to attain orbital velocity. For low Earth orbit, around 325 km in altitude, an object must travel about 28,000 km/hr in a direction perpendicular to the pull of gravity. How did we calculate this velocity?
Sir Isaac Newton provided the answer. He showed that: v = (GM/r)^{½}
Remember that the exponent of ½ means to take the square root of the term inside the parentheses.
Notice that this velocity is independent of the mass of the satellite and only depends on the mass of the central body (Earth in this case) and the distance from the center of the Earth, r.
The average radius of the Earth at its equator is 6,378.15 km. Let’s add 300 km to this for an orbiting altitude, so r = 6,378 + 300 = 6,678 km. The mass of the Earth is 5.974 x 10^{24} kg. Converting to kg-m-s units, we get the following equation for v:
v = (6.673 x 10^{-11} Nm^{2}/kg^{2 }x 5.974 x 10^{24} kg/ 6.678 x 10^{6}m)^{½} = 7.726 km/s = 27,815 km/hr
Try This!
Calculate orbital velocities for the following bodies:
Object | Mass (kg) | Radius (km) | Orbital Radius (km) | Orbital Velocity (km/s) |
Moon | 7.36 x 10^{22} | 1,737 | 1,937 | |
Mars | 6.42 x 10^{23} | 3,397 | 3,597 | |
Sun | 2.00 x 10^{30} | 695,500 | 149.6 | |
Neutron Star | 2.8 x 10^{30} | 20 | 220 | |
Milky Way Galaxy | 1.16 x 10^{42} | 2.55 x 10^{17} | 2.55 x 10^{17} | |
Basketball | 0.5 | 4.7 x 10^{-5} | 0.001 |
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