In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.
The weight of the object is
(mass) x (gravity in the place where the object is) .
On the surface of the Earth,
Weight = (60 kg) x (9.8 m/s²)
= 588 Newtons.
Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is
(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.
The object's weight would also be 0.04 of its weight on the surface.
(0.04) x (588 Newtons) = 23.52 Newtons.
Again, the object's mass is still 60 kg out there.
___________________________________________
If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
Imma go with A.
Hope this helps:)
I might be wrong but I'm pretty sure It's B
