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Assume that because of in creasing $\mathrm{CO}_{2}$ concentration in the atmosphere, the net radiation energy transfer rate for Earth and its atmosphere is $+0.002 \times 1350 \mathrm{W} / \mathrm{m}^{2}$ ,corresponding to a 0.2% decrease in radiation leaving Earth. (a) Determine the extra thermal energy added to Earth and its atmosphere in 10 years. [Note:The radiation falls on an area approximately equal to $\pi r_{\text { Earth }}^{2}$ where $r_{\text { Earth }}=6.4 \times 10^{6} \mathrm{m.} ]$ (b) If 30% of this energy is used to melt the polar ice caps, how many kilograms of ice will melt in 10 years? (c) How many cubic meters of ice will melt? (d) By how much will the level of the oceans rise in 10 years?

The energy from solar radiation falling on Earth is

1.07 \times 10 ^ { 16 } \mathrm { kJ } / \mathrm { min }

How much loss of mass from the Sun occurs in one day from just the energy falling on Earth?

The orbital velocity of Earth about the Sun is $30 \mathrm{~km} / \mathrm{s}$ . If Earth were suddenly stopped in its tracks, it would simply fall radially into the Sun. A congressional leader suggests firing a rocket loaded with radioactive wastes into the Sun for permanent disposal. For this to occur, with what speed and in what direction with respect to Earth's orbit should the rocket be fired?

Question

The radius of Earth is 6370 km, and its orbital speed about the Sun is 30 km/s. Suppose Earth moves past an observer at this speed. To the observer, by how much does Earth’s diameter contract along the direction of motion?

Solution

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Step 1

1 of 2

Let $D_0$ be the diameter of the earth at rest and $D$ is the diameter of the earth to the observer, the magnitude of the contraction is given by:

\begin{align}|\Delta D|=D_0-D\end{align}

the diameter of the earth to the observer is given by:

\begin{align} D=\dfrac{D_0}{\gamma}\end{align}

where $\gamma$ is the Lorentz factor and it is given by:

\gamma=\dfrac{1}{\sqrt{1-(v/c)^2}}

where $v$ is the radial speed. Substitute into (2) then into (1) to get:

D=D_0\sqrt{1-(v/c)^2}

thus,

|\Delta D|=D_0-D_0\sqrt{1-(v/c)^2}=D_0\left(1-\sqrt{1-(v/c)^2} \right)

substitute with the givens to get (note that $D_0=2 \times 6370 \times 10^{3} \mathrm{~m}=1.274\times 10^{7}$ m):

\begin{align*}|\Delta D|&= (1.274\times 10^{7} \mathrm{~m})\left(1-\sqrt{1-((30 \times 10^{3} \mathrm{~m/s})/(3.0 \times 10^{8} \mathrm{~m/s}))^2} \right)\\ &=0.0637 \mathrm{~m} \end{align*}

\boxed{|\Delta D|=0.0637 \mathrm{~m}}

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