Posted by leonard on Thursday, October 17, 2013 at 10:19am.
sqrt(( (2*sqrt(2)-2)*k*R^2)/m+2*g*R)
(2*(sqrt(2)-1)*k*R)+(2*m*g)-((sqrt(2)-1)*(k*R)*(1/sqrt(2)))
A satellite with a mass of ms = 8.00 × 103 kg is in a planet's equatorial plane in a circular "synchronous" orbit. This means that an observer at the equator will see the satellite being stationary overhead (see figure below). The planet has mass mp = 9.59 × 1025 kg and a day of length T = 0.5 earth days (1 earth day = 24 hours).
(a) How far from the center (in m) of the planet is the satellite?
(b) What is the escape velocity (in km/sec) of any object that is at the same distance from the center of the planet that you calculated in (a)?
A bungee jumper jumps (with no initial speed) from a tall bridge attached to a light elastic cord (bungee cord) of unstretched length L. The cord first straightens and then extends as the jumper falls. This prevents her from hitting the water! Suppose that the bungee cord behaves like a spring with spring constant k = 110 N/m. The bridge is h = 110 m high and the jumper's mass is m = 50 kg. Use g = 10 m/s2.
(a) What is the maximum allowed length L of the unstretched bungee cord (in m) to keep the jumper alive? (Assume that the spring constant doesn't depend on L).
(b) Before jumping, our jumper verified the spring constant of the cord. She lowered herself very slowly from the bridge to the full extent of the cord and when she is at rest she measured the distance to the water surface. What was the measured distance (in m)?
A planet has a single moon that is solely influenced by the gravitational interaction between the two bodies. We will assume that the moon is moving in a circular orbit around the planet and that the moon travels with a constant speed in that orbit. The mass of the planet is mp = 5.69 × 1025 kg. The mass of the moon is mm = 8.25 × 1022 kg. The radius of the orbit is R = 7.76 × 108 m.
1)What is the period of the moon's orbit around the planet in earth days (1 earth day = 24 hours).
help smeone plzz.!!