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?
November 7, 2009
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.
November 7, 2009
ya know you would probably have to ask someone who knows a bit better math…haha well that is a good question though…
November 7, 2009
…Superman is not encumbered with “our” Natural and Universal Laws… that’s why “he’s” Superman….
November 7, 2009
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!
November 7, 2009
yes absolutely right minimum speed11km/s
November 7, 2009
you really have watched superman a lot haven’t you? you researched a lot too…from the looks of it…
Wow…
November 7, 2009
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.
November 7, 2009
Escape Velocity: 11.186 km/s
Note: Not Speed but Velocity.
November 7, 2009
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.
November 7, 2009
To Escape graviataional pull , one must attain speed greater than 11.2km per second
November 7, 2009
You’ve basically said it yourself (or the homework question did?)
0.5mv^2 >= GmM/r
so the m’s (Superman’s mass) cancel out leaving:
0.5v^2>=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 >= 11.18 km/s.
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