Earth Radius Shrunk to 4.13 x 10^12 for 50 km.h-1 Escape Speed

[warmfire540]

To what radius would the Earth have to be shrunk, with no loss of mass, so that the escape speed from the surface of the Earth is 50 km.h-1

50kmh-1 = 13.9 m/s

v=sqrt(2GM/r)
G=6.67 x 10^-11
M=5.98 x 10^24
r=?

13.9=sqrt(7.98 x 10^14/r)
193.21=7.98 x 10^14/r
193.21r=7.98 x 10^14
r=7.98 x 10^14/193.21
r=4.13 x 10^12

 

[G01]

Looks good to me. Any specific question?

 

[alphysicist]

Hi warmfire540,

Are you sure you copied the problem down correctly? I can see where it doesn't seem right to you--the problem indicates that the radius you solve for should be smaller than the Earth's radius, but the one you calculate is much larger.

Did you copy the given escape speed correctly? Could it have supposed to have been something like 50000 km/hour? The real escape speed of the Earth is around 40000 km/hour, and if we want the radius to be smaller we need the given speed to be larger.