Ida u a small asteroid which lies in the asteroid belt, between Mars and Jupiter

Question

Ida u a small asteroid which lies in the asteroid belt, between Mars and Jupiter. Ida is roughly rectangular, around 35 miles long and 15 miles wide. (A nice photo is posted on the website.) To make things easier, let's calculate as if Ida is a sphere with radius r - 3 times 10 2 m (- 20 miles) and mass of 1 times 10 18 kg. Compute the value of g at the surface of Ida. (Can you derive the g equation?) On Earth, let's say you have a vertical leap of 0.50 m. Compute your launch velocity" on Earth. If you had the same launch velocity on Ida. how high could you jump? What would be your hang time on Ida? If you were standing on Ida. you could practically throw a ball into orbit! Compute the orbital velocity of an object just above the surface of Ida. (Can you derive the orbital velocity equation?) In 1993. space probe Galileo flew to within 1500 miles of Ida and took the photo that appears on our website. In those photos it was discovered that approximately 100 km from Ida was a tiny piece of rock barely one mile in diameter, subsequently named Dactyl. It turned out that Dactyl was in orbit around Ida. That was the first -- and I think still the only -- known case of an asteroid having a satellite (a "moonlet", astronomers call it). Compute the orbital speed of Dactyl, calculating as if it travels in a circular orbit of radius 100 Ian.

Explanation / Answer

yo yo, the g equation's like this. Fg=GMm/r^2 mg=GMm/r^2 g=GM/r^2 g=(6.67x10^-11)(10^18)/ (3x10^4)^2=.07411m/s^2 On earth, if you wanted to jump.5m then to have your potential energy equal your kinetic energy mgh=1/2 mv^2 2g(.5M)=v^2 v=sqrt(2g(.5M)) v=3.13m/s with the same launch velocity on Ida mgh=1/2mv^2 h=(1/2v^2)/g=.5(3.13^2)/(.07411)=66.12M You'd be in the air x=1/2at^2 => t=sqrt(2x/a)=sqrt(2x66.12M/.07411)=1784s one way. So in total 3569s For a ball to orbit, it's centripetal acceleration must equal the acc of gravity v^2/r= .07411m/s^2 v=sqrt((3x10^4)(.07411))=47.15m/s For a moonlet to orbit, it's cetripetal acceleration must equal the acceleration due to gravity V^2/r=GM/r^2 V^2=GM/r V=sqrt(6.67x10^-11 x 10^18 / 100000)=25.8m/s

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