What speed must Superman attain in order to escape from the Earth's gravitational pull?

Superman (99 kg) is often seen to fly up and away from the Earth's atmosphere. In order to accomplish this feat, Superman's kinetic energy (0.5mv2) must at least be equal to the gravitational potential energy at the Earth's surface (GmM/r). If the Earth's radius is 6,378,100 meters and its mass is 5.9742 x 1024 kg, calculate the speed Superman must attain in order to escape from the Earth's gravitational pull. Is this speed dependent on Superman's mass?|||The first answer is technically correct. However, since Superman can violate the laws of gravity, it would not be unreasonable to assume that he could simply climb out of earth's gravity well at any speed he damned well pleases.|||great answer bro!!

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|||All I knew is the 'Superman' could move faster than the fastest bullet at that time...and he also caused our world lunar calendar going wrong...hints:he changed the orbital of the moon......[superman 4]


'Could we standardized our new lunar calendar' back?

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|||and by the way...remember..earlier than that..he already changed the earth time to slightly an hour or more...by using his powers to make the reversal earth orbital..[-time]...some how he is stronger than been told..probably he could overcome the forces of gravity..and hints:E=mcxc [a time travel?]

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|||Oh come on with all that everyone knows he's faster than a speeding bullet, more powerful than a locomotive, and able to leap tall buildings with a single bound, so who needs figures lol!|||ya know you would probably have to ask someone who knows a bit better math...haha well that is a good question though...|||To Escape graviataional pull , one must attain speed greater than 11.2km per second|||yes absolutely right minimum speed11km/s|||Escape Velocity: 11.186 km/s





Note: Not Speed but Velocity.|||Yeah 11 km per sec is the required speed. BUt people do easily get confused with the reqired velocit and the applied force. So if superman puts on or lose weight still he needs to attain 11km /sec but he shall be using different amount of energies.|||You've basically said it yourself (or the homework question did?)


0.5mv^2 %26gt;= GmM/r





so the m's (Superman's mass) cancel out leaving:


0.5v^2%26gt;=GM/r





Thus the speed (escape velocity) doesn't depend on Superman's (or a ball or a satellite, etc.) mass.





The gravitational constant G is 6.674 x 10^-11 m^3 kg^-1 s^-2 but for the Earth (and the Sun and other planets) the product GM is known with better accuracy and is 398600.44 km^3 s^-2 for the Earth. Divide by Earth's radius of 6378.1 km gives 62.495 km^2 s^-2. Multiply by 2 and then take the square root gives:


v %26gt;= 11.18 km/s.|||...Superman is not encumbered with "our" Natural and Universal Laws... that's why "he's" Superman....|||The escape velocity for Earth is about 11km/sec. It is independent of the escape object's mass. However, the more massive the escape object, the more force is required to accelerate that object to the required 11km/sec.





By the way, Superman has never left Earth's gravitation field. Remeber that the moon still orbits the Earth, which means the moon is still in Earth's gravitational field. I don't recall superman ever going beyond the Moon's orbit.|||you really have watched superman a lot haven't you? you researched a lot too...from the looks of it...


Wow...