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Calculate the daily loss of energy by th...

Calculate the daily loss of energy by the earth, if the temperature gradient in the earth's crust is `32^(@)C` per km and mean conductivity of the rock is `0.008` of CGS unit. (Given radius of earth `=6400km`)

A

`10^(40)cal`

B

`10^(30)cal`

C

`10^(18)cal`

D

`10^(10)cal`

Text Solution

Verified by Experts

(c) Temperature gradient `(d theta)/(dx)=32/1000.^(@)C//m=32/(10^(5)).^(@)C//m`
Loss of energy by earth is given by
`Q=KA(d theta)/(dx)xx86400`
`=0.008xx(4xx22)/7xx(64xx10^(8))^(2)xx32/(10^(5))xx86400`
`=1.1xx10^(18)cal=10^(18)`cal
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Knowledge Check

  • The radius of the earth is 6400 km, the order of magnitude is

    A
    `10^(7)`m
    B
    `10^(4)`m
    C
    `10^(3)`m
    D
    `10^(2)m`

  • Find the velocity of a satellite at height 80 km from earth. If the radius of earth is 6400 km

    A
    7 km/s
    B
    8 km/s
    C
    7.84 km/s
    D
    11.2 km/s

  • Calculate the acceleration due to gravity at a height of 1600 km from the surface of the Earth. (Given acceleration due to gravity on the surface of the Earth g_(0) = 9.8 ms^(-2) and radius of earth, R = 6400 km).

    A
    `6.27 m//s^2`
    B
    `3.28 m//s^2`
    C
    `5.36 m//s^2`
    D
    `4.86 m//s^2`

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