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---------------------------- - 2247 - How Fast to Orbit?
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- Speed is what it takes to go into orbit. It does not matter how big you are, just how fast you are going. This Review does the calculations.
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- Gravity Probe B is a satellite in orbit 400 miles up. ( See Review # 1536 to learn the Gravity Probe B mission). How fast must the satellite be going in order to stay in this orbit. A satellite in orbit is weightless because it is continuously falling around the Earth. Its orbit is falling around the curvature of the Earth. The rate of falling is determined by the acceleration of gravity, not the mass of the satellite. It is a constant. The rate of falling is the same for every mass that is in free fall and is at the same distance from the center of Earth, or, the center of gravity. The acceleration of gravity near the Earth is 32.2 feet per second per second. (that is 9.8 m/sec^2)
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- The satellite , like the Moon, and the Planets obey Kepler’s laws of motion. The cube of the radius of orbit is proportional to the square of the period of orbit. Let’s start with the Moon, that is also in free fall around the Earth:
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------------------ The period of the Moon’s orbit is 27.32 days.
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------------------ The radius of the Moon’s orbit is 239,509 miles.
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- The cube of the radius is proportional to the square of the period:
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- ------------------ (239,509 )^3 = K* ( 27.32 days)^2
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-------------------- 13/57 *10^15 = K * (7.464*10^2)
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- “K” is the constant of proportionality depending on the units of measurement used, and, we do not have to know that to solve the period for the satellite. We know that regardless of mass the ratios are the same.
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------------------- Satellite Radius^3 / Moon Radius^3 // Time, Ts^2 / Moon Period^2
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-------------------- Period for the satellite = Ts
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-------------------- Radius of orbit for the satellite is 400 miles plus the radius of the Earth 3957 miles equals 4,357 miles.
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- Now equate the two ratios and solve for Ts:
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-------------------- (4.357*10^3)^3 / 2.39509 *10^5)^3 = Ts^2 / (27.32)^2
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-------------------- 6.02*10^-6 = Ts^2 / 7.46*10^2
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------------------- Ts^2 = 44.93*10-4
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-------------------- Ts = 6.7 *10*-2 days
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-------------------- Ts = 1.61 hours ( 96.6 minutes )
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- The time for one complete orbit of the satellite is 1.61 hours. Now if we know the distance traveled in one complete orbit we can calculate the speed.
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- The distance of the orbit is the circumference of a circle which is 2 * pi * r. The velocity of orbit is distance / time.
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------------------ v = d / t = 2*pi *r / t = 2*pi*(4,357 miles) / 1.61 hours.
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------------------- v = 17,004 miles per hour.
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- However, this assumes we have a perfect circle for an orbit. What would be the effect if the satellite had a slightly elliptical orbit? An elliptical orbit requires that it travel a greater amount of distance in the same amount of time, so it must be going faster during part of the orbit. Can we do this calculation without assuming the perfect circle orbit?
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- With the elliptical orbit the radius of orbit,” r“, changes with the function of time, t. If we construct a coordinate plane with the radius connected from the center of the Earth to the point on the orbit , we can make the radius to to be hypotenuse of a right triangle.
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------------------ hypotenuse = r
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----------------- opposite side = r * sin a, where “ a “ is the angle
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----------------- adjacent side = r cos a
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----------------- Using the Pythagorean Theorem: r^2 = (r*sin a)^2 + (r*cos a)^2
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- The radius is rotating as the points move around the elliptical orbit. The radius is a function of time, r(t), depending on the angularly velocity of rotation. The angular velocity = da / dt. The rate of change of the angle with time. Restating the distance equation in terms of changes with time:
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--------------------- r(t)^2 = (r * sin da /dt * t )^2 + ( r*cos da/dt * t)^2
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- The speed of the satellite is constant, but, the velocity of the satellite is constantly changing, because the direction of travel is constantly changing. Velocity is the rate of change of distance. v(t) = dr /dt. If we take the derivative of both sides of the above equation:
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---------------- v(t)^2 = ( dr/dt)^2 = r^2 cos a^2 (dc/dt)^2 + r^2 *(- sin a)^2 ( dc/dt)^2
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- Where: the derivative of sin a = cos a dt
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- Where: the derivative of cos a = - sin a dt
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- Speed is the absolute magnitude of velocity. So, the square of velocity gets us an absolute magnitude and is equivalent to speed.
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------------------------ (cos a)^2 + (sin a)^2 = 1
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- Again this is the Pythagorean theorem: opposite^2 / r^2 + adjacent^2 / r^2 = 1
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------------------- v(t)^2 = r^2 ( da/dt)^2
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------------------- v = r * da/dt
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------------------- velocity = radius * angular velocity
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------------------ (da/dt)^2 = v^2 / r^2
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- Acceleration is the rate of change of velocity. acceleration = dv / dt = r * (da/dt)^2
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------------------------------ acceleration = r * v^2 / r^2
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------------------------------ acceleration = v^2 / r
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----------------------------- In the case of a satellite in continuous orbit the acceleration of orbit must balance with the acceleration of gravity, g.
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-------------------------------acceleration = g = 32.2 feet/sec^2
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------------------------------ v^2 = g*r
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- The constant acceleration of gravity near the Earth is 32.2 feet per second per second, or 79,036 miles per hour ^2
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--------------------------- v^2 = 32.2 feet /sec^2 * 4.357*10^3 miles * mile / 5,280 feet * (3600 sec)^2 / hour ^2
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-------------------------- v^2 = 344.4 *10^6 miles^2 / hours^2
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------------------------- v = 18,557 miles per hour.
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- The assumption here is that the acceleration of gravity is 32.2 feet /sec^2 at 400 miles altitude. Assuming that is near to the Earth. The other assumption was that the orbit was a perfect circle. Both need some tweaking to get a more accurate answer to how fast the satellite is traveling. All velocity is relative depending on your assumptions. But, the satellite must be traveling about 18,000 miles per hour. It will be traveling slightly faster if its orbit gets closer to Earth and slightly slower when farther away from Earth.
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- Now you know “how fast to orbit”. Works for planets, satellites, and even galaxies.
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- January 25, 2019. 1780
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--------------------- Friday, January 25, 2019 -------------------------
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