A free fall experiment is performed on a newly discovered spherical planetoid wh

Question

A free fall experiment is performed on a newly discovered spherical planetoid where an 80kg astronaut dropped a rock from a height of 216cm and its falling time was electronically measured to be 1.2 secs. The planets circumference was determined to be 36.74*10^6 m by geometrical methods. The astronaut is fully aware that h = (1/2) g t^2. Determine the following; A) The weight of the astronaut on the planets surface B) The mass M of the planetoid C) The orbital speed , V orb, near the planetoids surface D) The orbital period , T, near the planetoids surface E) The altitude , h, for a geo-synchronous satellite orbiting the planetoid with a rotational period of 15 hrs

Explanation / Answer

A) Since h = (1/2)gt^2, we can plug in the values for h and t and solve for g. 0.216 = (1/2)g(1.2)^2 g = 2*0.216 / (1.2)^2 g = 0.30 m/s^2 The astronaut's weight is mg, so that is: (80)(0.30) = 24.0 N B) Radius of the planet is: 36.74*10^6 / (2*pi) = 5.847*10^6 Using the universal gravitation equation and the values we found in part (a), we can solve for the mass of the planet. F = Gm1m2 / r^2 24.0 N = (6.674*10^-11)(80) m2 / (5.847*10^6 )^2 m2 = 1.537 * 10^23 C) Vorb = sqrt(GM / r) where M is the mass of the planet Vorb = sqrt((6.674*10^-11)(1.537 * 10^23)/(5.847*10^6 )) Vorb = 1324.54 m/s D) Vorb = 2pi*r / T where T is the period 1324.54 = 2pi*(5.847*10^6 ) / T T=27736.3 s This is 7.7045 hours E) Assume that the velocity of the satellite is the same as the rotation of the planet. Now solve: Vorb = 2pi*r / T 1324.54 = 2pi*r / (15*3600) r = 1.1384 * 10^7 Subtract the radius of the planet from this value to get: h = 1.1384 * 10^7 - 5.847*10^6 = 5.537*10^6 m

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