- #1
SnowOwl18
- 71
- 0
Another problem (oy!):
------A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of about D = 1.02km. What must be the rotation speed (in revolutions per day) if an effect equal to gravity at the surface of the Earth (1g) is to be felt? Do not enter units. ---------
Ok, so first I found the velocity of the station by doing v= sqrt r x g . I got 70.69 m/s. And then I used the formula T= 2PiR/V ...and I got 45.33 rps...or at least I think it's revolutions per second. It asks for revolutions per day, so I multiply 3600s (per hour) by 24 hours (in a day) and get 86400...so I multiplied that by 45.33 to get 3.9e6...but apparently I'm wrong. If anyone could be so kind as to enlighten me as to what I'm doing wrong I'd greatly appreciate it. Thanks! (again).
------A projected space station consists of a circular tube that is set rotating about its center (like a tubular bicycle tire). The circle formed by the tube has a diameter of about D = 1.02km. What must be the rotation speed (in revolutions per day) if an effect equal to gravity at the surface of the Earth (1g) is to be felt? Do not enter units. ---------
Ok, so first I found the velocity of the station by doing v= sqrt r x g . I got 70.69 m/s. And then I used the formula T= 2PiR/V ...and I got 45.33 rps...or at least I think it's revolutions per second. It asks for revolutions per day, so I multiply 3600s (per hour) by 24 hours (in a day) and get 86400...so I multiplied that by 45.33 to get 3.9e6...but apparently I'm wrong. If anyone could be so kind as to enlighten me as to what I'm doing wrong I'd greatly appreciate it. Thanks! (again).