Answer:
The scale reading at the pole is higher. The difference is 8.019 N.
Explanation:
The centrifugal force of the equator is the same as the centrifugal force of the Earth itself. On the other hand, the centrifugal force on the poles are zero, since they are not spinning in the way the equator is. The centrifugal force on a body at the equator is 0.034 m/s^2 times the mass of the body. The centrifugal force at the poles is zero. Therefore, the total weight of the man at sea level at the equator (gravity minus centrifugal force) is 9.764 m/s^2 times his mass, whereas his weight is 9.863 m/s^2 times his mass at the poles. Using simple multiplication, I figured out his weight at the north pole is 798.903 N, and his weight at the equator is 790.884 N. The difference of these two numbers is 8.019 N.
Hope this helps!
Source: https://www.wtamu.edu/~cbaird/sq/2014/01/07/do-i-weigh-less-on-the-equator-than-at-the-north-pole/
The location where the weight of a person will be more is required.
The scale at the pole has a higher reading.
The difference in weight is 4.212 N.
m = Mass of man = 81 kg
The acceleration due to gravity on the poles is [tex]g_p=9.832\ \text{m/s}^2[/tex]
The acceleration due to gravity on the equator is [tex]g_e=9.78\ \text{m/s}^2[/tex]
Gravitational acceleration is given by
[tex]g=\dfrac{GM}{R^2}[/tex]
where,
M = Mass of Earth
G = Gravitational constant
R = Radius
The Earth in not a perfect sphere it is wider at the equator than at the poles.
As [tex]R[/tex] increases, the value of [tex]g[/tex] decreases.
Weight of the person at the north pole
[tex]w_p=mg_p\\\Rightarrow w_p=81\times 9.832=796.392\ \text{N}[/tex]
Weight of the person at the Equator
[tex]w_e=mg_e\\\Rightarrow w_e=81\times 9.78=792.18\ \text{N}[/tex]
The difference in the weight is [tex]796.392-792.18=4.212\ \text{N}[/tex]
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