Question

A 71 kg man weighs more at the north pole than at the equator. By how much?

Answer 1

update: tried 3.692 and it was incorrect

Answer 2

You would weigh about 0.3% less at the equator than at the poles.

Answer 3

weight of an object depends on the gravity. So we need to know what is gravity at North Pole and what is the gravity at Equator

Answer 4

On wikipedia it says gravity is 9.780 m/s^2 at the equator and 9.832 m/s^2 at the north pole

Answer 5

There is also a very minor effect of rotation - the body at the pole is not travelling in the circle around the equator.

Since the period is 24hours I would suspect this would be an almost insignificant effect....

Answer 6

then the weight is
70*9.832=...?Newtons at North pole
and
70*9.780=... Newtons at Equator

Answer 7

71 you mean?

Answer 8

yes! sorry!

Answer 9

I did both and subtracted the difference and got 3.962 N

Answer 10

hmmm - my rough calculation makes the rotation effect approx. 2N
so maybe significant

Answer 11

no, you have to compute the ratio, between the weight at north pole over the weight at Equator, like this:

\[\Large \frac{{71 \times 9.832}}{{71 \times 9.780}} = \frac{{9.832}}{{9.870}} = ...?\]

Answer 12

oops:
\[\Large \frac{{71 \times 9.832}}{{71 \times 9.780}} = \frac{{9.832}}{{9.780}} = ...?\]

Answer 13

@Michele_Laino

It asks for the difference in weight - not the ratio

subtract the 2 values....

Answer 14

yes! you are right! @MrNood

Answer 15

I got 3.962 and it's incorrect

Answer 16

Does someone mind checking my math?

Answer 17

your maths is correct if the figures are correct

Answer 18

How do you know your answer is incorrect? what value do you have for an answer?

Answer 19

its an online program that gives immediate feedback if I submit a wrong answer. It's in newtons

Answer 20

so you have no ide if it is just a rounding error - or minor changes in the values for g?

(see this quote:
we find that the force of earth’s gravity on your body at the equator is 9.798 m/s2 times the mass of your body, whereas at the poles it is 9.863 m/s2 times the mass of your body)

There is evidently some room for manouevre

Answer 21

THIS site ALSO suggest you need to account for the rotational forces, and ALSO calculates about 2N for that effect

https://sciencequestionswithchris.wordpress.com/2014/01/07/do-i-weigh-less-on-the-equator-than-at-the-north-pole/

Answer 22

it could be. I'll try the same thing with those values for g.

Answer 23

With those new values for g I got 4.615 and it's still incorrect

Answer 24

Are you saying I should subtract 2N?

Answer 25

both effect act to reduce your weight at the equator

so if the g effecct is 4N and the rotation effect is 2 N then the overall effect would be 6N less at the equator.

(the site above quotes 1% - which would be 7N for the 71 kg body)

Answer 26

I think you will struggle to get exactly the right answer - If it is online then it probably only going to accept your answer if it EXACTLY matches the programmed answer.
You do not know what precision to enter it, or the values to take for g.

I think you have shown you understand the concept so unless it is a compulsory test I should move on. Not all questions are 'good questions' for the sake of learning

Answer 27

you need to calculate the gravity at the north pole first
Fg= ( G m1 M2 )/ r^2

G :universal gravitation constant = 6.673 x 10^(-11) N m2/kg2
m1: mass of the earth
m2: body mass ( man) but this mass also too small compare to the earth so it could be ignored
r: earth radius

Answer 28

Equatorial radius (km) 6378.1
Polar radius (km) 6356.8

Answer 29

Core radius (km) 3485

Answer 30

You cannot ignore m2

THAT is the mass of the body 71

The force is directly proportional to THAT

However - you will still come up with 'debatable answers, and unless she is given the values ot use , and the precision to answer, then she is unlikely ot exactly match the 'computer' answer

Answer 31

you have quoted radius to the nearest 100m - where was that measured - to the top of the hill? or the bottom of the valley?