Weight decrease of 1% at 45° latitude with elevation

[shaiqbashir]

Hi!

Im getting confused in solving this problem. I just can't pick out the correct formulae to solve it.

"At 45 degree latitude, the gravitational acceleration as a function of elevation "z" above sea level is given by

g=a-bz

where a=9.807 meter per second square and b=3.32x10^-6 s^-2.

Determine the height above the sea level where the weight of an object will decrease by 1 percent."

Thanks in advance

 

[dextercioby]

What is the value of "g",the one dropped by 1%...?And what is the height...?

Daniel.

 

[shaiqbashir]

My dear friend!

The height u r required to determine!

 

[dextercioby]

Well,that wasn't the question... Did u computethe new "g"...?

Daniel.

 

[snbose]

Solution:
At z=0 (sea level); g=a
Say at height x the accelarion due to gravity is 99% i.e 0.99a. This is the height where the weight will be 99% too (F=mg).
So 0.99a=a - bx.

x = 0.01a/b.

 

[dextercioby]

It would have been much better,if the OP would have found the sollution...

Daniel.

 

[shaiqbashir]

Thanks a lot Snbose! You have helped me a lot

Thank u very much