Physics: Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite distance apart. Because of their gravitational attraction, they head toward each other on a collision course. (a) When their center-to-center separation is d, find expressions for the speed of each planet and their relative velocity. (b) Find the kinetic energy of each planet just before they collide, if m1 2.00 1024 kg, m2 8.00 1024 kg, r1 3.00 106 m, and r 2 5.00 106 m.

The emissivity of the radiation shield is Explanation:From the question we are told that The temperature of the first parallel plate is The temperature of the second parallel plate is The emissivity of first plate is The emissivity of first plate is Generally the total radiation heat that is been transferred without the shield is mathematically represented as Where is the Stefan-Boltzmann constant which has a value Substituting values ...

Two hypothetical planets of masses m1 and m2 and radii

Ответ:

ariellllm

22.01.2022

Physics

(a) $v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) Kinetic Energy of planet with mass m₁, is KE₁ = 1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ = 2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

$F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}$

Energy is given by

$Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ = m₁·v₁/m₂

At equilibrium, we have;

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$ which gives

$2G{m_{1}m_{2}}= d m_{1} v^2_1+ dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)$

multiplying both sides by m₂/m₁, we have

$2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)$

Such that v₁ = $\sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }$

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

Similarly, with v₁ = m₂·v₂/m₁, we have

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow 2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)$

From which we have;

$2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2$ and

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

The relative velocity = v₁ + v₂ = $v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

v₁ + v₂ = $(m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) The kinetic energy KE = $\frac{1}{2}mv^2$

$KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2$

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

$v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ = 10333.696 m/s

$v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² = 1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² = 2.6696×10³¹ J.

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Ответ:

tiakimberly3

04.10.2020

Physics

The emissivity of the radiation shield is e_3 = 0.580

Explanation:

From the question we are told that

The temperature of the first parallel plate is T_1 = 650K

The temperature of the second parallel plate is T_2 = 400K

The emissivity of first plate is e_1 = 0.6

The emissivity of first plate is e_2 = 0.9

Generally the total radiation heat that is been transferred without the shield is mathematically represented as

$Q_1 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{\frac{1}{e1 } + \frac{1}{e_2} -1 }$

Where $\sigma$ is the Stefan-Boltzmann constant which has a value $5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}$

Substituting values

$Q_1 = \frac{ 5.67 *10^{-8} (650 ^4 - 400 ^4)}{\frac{1}{0.6 } + \frac{1}{0.9} -1 }$

$Q_1 = 4876.8 \ W/m^2$

From the question we are told the that using the radiation shield would reduce the radiation heat transfer by 15%

So the new heat transfer is

$Q_2 = \frac{15}{100} * Q_1$

So $Q_2 = \frac{15}{100} * 4876.8$

Q_2 = 731.52 W/m^2

Now this new radiation heat transfer can be mathematically represented as

$Q_2 = \frac{\sigma (T_1 ^4 - T_2 ^4)}{ [\frac{1}{e_1 } + \frac{1}{e_2 } - 1 ] + n [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

Where e_3 the emissivity of the radiation shield and n is the number of radiation shield

Substituting values

$731.52 = \frac{5.67 *10^{-8} (650 ^4 - 400 ^4)}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$731.52 = \frac{1.4175*10^{-5}}{ [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ]}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = \frac{1.4175*10^{-5}}{731.52}$

$[\frac{1}{0.6} + \frac{1}{0.9 } - 1 ] + 1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8}$

$1 [\frac{1}{e_3} + \frac{1}{e_3} -1 ] = 1.9377*10^{-8} - [\frac{1}{0.6} + \frac{1}{0.9 } - 1 ]$

$\frac{1}{e_3} + \frac{1}{e_3} = 1 .7222222416$

e_3 = 0.580

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