Name: Joshua D.
ProgramYear: 2006
Submit Date: May 28, 2006
Review Date: Jun 16, 2006

(ALL ANSWERS GIVEN IN SAME SIGNIFICANT DIGITS AS PROVIDED FROM PROBLEM DATA, ALLOWING FOR STATISICAL DEVIATION) 1. Compare the Pathfinder velocity at a with earth's orbital velocity at a. What is the difference and why? (Express your answer in terms of total orbital energy.) At point (a), Earth's orbital velocity is about 30 km/sec or 67,000 mph and the Pathfinder's velocity is about 32 km/sec or 72,000 mph. This difference of 2 km/sec or 5,000 mph is because the pathfinder and the Earth are now in different orbits. The pathfinder (now in its Hohmann Transfer Orbit) is heading in a much different direction than the earth once it escapes Earth's immediate gravitational pull. The different orbit requires a much faster speed at which the Pathfinder must travel. The total orbital energy (gravitational potential energy and kinetic energy) of the orbits now vary. While the gravitational potential energy of both the Pathfinder and the Earth remain the same (due to their equal distance from the Sun), the Pathfinder's kinetic energy must differ because of its now separate orbit. Its kinetic energy, and thus, its orbital velocity, are now much greater. ///MATH WORK\\ Given: ---EARTH VELOCITY--- * Distance from Earth to Sun = 150,000,000 km * Earth Orbital Period = 365.25 days * Orbital Velocity = (2 * pi * distance) / period ---PATHFINDER VELOCITY--- * a = 190 Mkm = 1.9E11 m * Distance from the Earth to the Sun = 150 Mkm = 1.5E11 m * Velocity = 1.6 x 10^10 * Sqrt[1/r - 1/(2a)] Calculate: * Orbital Velocity of Earth * Velocity at point A Orbital Velocity = (2 * pi * distance) / period = (2 * 3.1416 * 150,000,000 km) / 365.25 days = About 2,580,370 km/day 2,580,370 km 1 day 1 hour 1 minute -------------- * ------------- * ------------- * ------------- = 29.87 km/sec 1 day 24 hours 60 minutes 60 seconds 2,580,370 km 1 day 1 mile -------------- * ------------- * ----------- = About 66,807 mph 1 day 24 hours 1.6093 km Velocity at point A = 1.6E10 * Sqrt[1/(1.5E11m) - 1/(2*1.9E11m)] = 3.2 * 10^4 m/s = 32 km/s = about 72000 mph 2. Compare the Pathfinder velocity at b with mars' orbital velocity at b. Again, what is the difference and why? (Express your answer in terms of total orbital energy.) At point (b), Mars' orbital velocity is about 24 km/sec or 54,000 mph and the Pathfinder's velocity is about 21 km/sec or 47,000 mph. This difference of 3 km/sec or 7,000 mph is because the pathfinder and the Mars are again in different orbits. The pathfinder (in its Hohmann Transfer Orbit) is meeting up with Mars' orbit and must go slower in order to be caught by Mars' gravitational pull and be locked into Mars orbit. The different orbit requires a much slower speed at which the Pathfinder must travel. The total orbital energy (gravitational potential energy and kinetic energy) of the orbits vary. While the gravitational potential energy of both the Pathfinder and Mars remain the same (due to their equal distance from the Sun), the Pathfinder's kinetic energy must differ because of its separate orbit. Its kinetic energy, and thus, its orbital velocity, are slower because of the stage of orbit that it is in. ///MATH WORK\\ Given: ---MARS VELOCITY--- * Distance from Mars to Sun = 230,000,000 km * Mars Orbital Period = 687 days * Orbital Velocity = (2 * pi * radius) / period ---PATHFINDER VELOCITY--- * a = 190 Mkm = 1.9E11 m * Distance from Mars to the Sun = 230 Mkm = 2.3E11 m * v = 1.6 x 10^10 * Sqrt[1/r - 1/(2a)] Calculate: * Orbital Velocity of Mars * Velocity at point B Orbital Velocity = (2 * pi * radius) / period = (2 * 3.1416 * 230,000,000 km) / 687 days = About 2,103,546 km/day 2,103,546 km 1 day 1 hour 1 minute -------------- * ------------- * ------------- * ------------- = 24.35 km/sec 1 day 24 hours 60 minutes 60 seconds 2,103,546 km 1 day 1 mile -------------- * ------------- * ----------- = About 54,462 mph 1 day 24 hours 1.6093 km Velocity at point B = 1.6E10 * Sqrt[1/(2.3E11m) - 1/(2*1.9E11m)] = 2.1 * 10^4 m/s = 21 km/s = about 47,000 mph 3. The Earth has a radius of 6400 km and spins once on its axis in 24 hours. Calculate the velocity of a point at the equator in km/sec and mph. The velocity of a point at the equator is 0.47 km/sec or 1000 mph. ///MATH WORK\\ Given: * Earth's radius = 6400 km * Earth's Axial Spin = 24 hrs Calculate: * Velocity of a point at equator Earth's Circumference = 2 * pi * radius = 2 * pi * 6400 km = 4.0 * 10^4 km 4.0 * 10^4 km 1 hour 1 minute -------------- * ------------- * ------------- = 0.47 km/sec 24 hours 60 minutes 60 seconds 4.0 * 10^4 km 1 mile -------------- * ----------- = 1000 mph 24 hours 1.6093 km 4. When viewed from celestial north, the Earth both rotates and revolves counter-clockwise. Do the orbital and rotational velocities add or subtract at local midnight? How about at local noon? What considerations might affect the time of day for a launch? Why did NASA launch the Pathfinder spacecraft eastward? At local midnight, the Earth's orbital and rotational velocities contribute to each other. A point on the Earth at this time (facing directly opposite the Sun), spinning and rotating counter-clockwise, will be traveling eastward both by its orbit and by its rotation. In contrast, at local noon, a point on the Earth, viewed from celestial north, will be traveling westward due to Earth's rotation. However, Earth's orbit is still moving eastward. Therefore, at noon, the orbital and rotational velocities are opposite and subtract from each other. Because of this difference, scientists may consider time as a factor when launching a spacecraft. If the launch uses Earth's counter-clockwise orbit as a contributing factor to the launch, scientists may consider also launching closer to midnight, gaining an extra eastward boost from Earth's rotation. Accordingly, if the spacecraft needs to travel in a celestially westward direction, a launch may be scheduled closer to noon, allowing for Earth's rotation to help the westward push. NASA most likely launched the Pathfinder eastward due to this rotational velocity. But launching eastward, the Pathfinder was given an extra boost from its already-traveling eastward direction from Earth's rotation. Thus, the eastward launch, combined with the eastward rotation, gave an even greater speed at which to enter the Hohmann Transfer Orbit. 5. In spacecraft design, energy is sometimes expressed in terms of change in velocity required to achieve orbit ( "delta-vee" or Dv). Given what we've just done, what Dv does the Pathfinder require at a? The Pathfinder's velocity at launch (near local midnight) would have been the speed of Earth's orbital velocity plus its rotational velocity (30 km/sec or 67,000 mph + 0.47 km/sec or 1000 mph, equaling 30.47km/sec or 68,000 mph). To achieve orbit, the Pathfinder must have a speed of 32 km/sec or 72,000 mph, given from the calculation of its speed at point A. Therefore, the Pathfinder needs to have an extra 1.6 km/sec or 4,000 mph boost in velocity. 6. Actually, additional energy (velocity) is required for a spacecraft just to escape the Earth's gravitational field. This velocity is given by the expression vEscape = (2GMEarth/rEarth)1/2. With MEarth = 6 X 1024 kg, calculate this velocity in km/sec and mph. This velocity must be added to the Dv calculated in Problem 5. How much, as a percent, does the result change compared to the value obtained in Problem 5? Does leaving the Earth's gravity well cost a lot in fuel? The new value, with an additional 11 km/sec is 12.6 km/sec. This is about an extra 700% increase in speed. Yes, leaving Earth's gravitational pull costs a lot of fuel. ///MATH WORK\\ Given: * Escape Velocity = sqrt(2GM/r) * Gravitational Constant = 6.6732 x 10^-11 m^3 / kg s^2 * Mass of the Earth = 6 x 10^24 kg * Radius of the Earth = 6.4 x 10^6 m Calculate: * Escape velocity of the Earth 2 * 6.6732 x 10^-11 m^3 / kg s^2 * 6 x 10^24 kg V_esc = sqrt( ------------------------------------------------------ ) 6.4 x 10^6 m 8 x 10^14 m^3 / s^2 V_esc = sqrt( ------------------------- ) 6.4 x 10^6 m V_esc = sqrt( 1 x 10^8 m^2 / s^2 ) V_esc = 1.12 x 10^4 m/s V_esc = 11 km/s Sources: * http://en.wikipedia.org/wiki/Orbital_velocity * http://en.wikipedia.org/wiki/Escape_velocity * Halliday, Resnick, and Walker. Fundamentals of Physics Extended, Fifth Edition. John Wiley & Sons, Inc. 1997.