Question

A body stretches a spring by a particular length at the earth's surface at equator. At what height above the south pole wil ilt stretch the same pering by the same length ? Assume the earth to be spherical.

Answer 1

Answer:

Height of the mass from the surface of earth must be equal to

$h = 1.1 \times 10^4 m$

Explanation:

As we know that the acceleration due to gravity at equator position is given as

$g' = g - \omega^2 R$

So we have

$m(g - \omega^2 R) = kx$

now at south pole there is no effect of rotation

so we have

$mg' = kx$

$mg' = m(g - \omega^2R)$

$g' = g - \omega^2R$

as we know that acceleration due to gravity at some height above the surface of earth is given as

$g = g(\frac{R}{R + h})^2$

so we have

$g(\frac{R}{R + h})^2 = g - \omega^2 R$

$g(1 - \frac{2h}{R}) = g - \omega^2 R$

$\frac{2gh}{R} = \omega^2 R$

$h = \frac{\omega^2 R^2}{2g}$

so we have

$h = \frac{2\pi^2 R^2}{gT^2}$

$h = \frac{2\pi^2(6.36 \times 10^6)^2}{9.81(24 \times 3600)^2}$

$h = 1.1 \times 10^4 m$

#Learn

Topic : Gravitation

https://brainly.in/question/13079049

Answer 2

Answer:

At equator g=g−ω

2

R...........(1)

Let at h height above the south pole,the acceleration due to gravity is same

Then,here g=(1−

R

2h

)...........(2)

g−ω

2

R=g(1−

R

2h

)

or1−

g

ω

2

R

=1−

R

2h

or h=

2g

ω

2

R

2

=

2×9.81

(7.3×10

−5

)

2

×(6400×10

3

)

2

=11125N=10Km(approximately)