Answer:
v₂ = 7.6 x 10⁴ m/s
Explanation:
given,
speed of comet(v₁) = 1.6 x 10⁴ m/s
distance (d₁)= 2.7 x 10¹¹ m
to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m
v₂ = ?
speed of planet can be determine using conservation of energy
K.E₁ + P.E₁ = K.E₂ + P.E₂
[tex]\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}[/tex]
[tex]\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}[/tex]
[tex]v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}[/tex]
[tex]v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}[/tex]
[tex]v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}[/tex]
v₂ = 7.6 x 10⁴ m/s
The speed when at a distance of [tex]4.8*10^{10} m[/tex] is mathematically given as
vf = 6.92 * 10^(4) m/s
Generally, the equation for the conservation of energy is mathematically given as
E = (1/2)mv^2 - GmM/r
Where
E_i = E_f
Hence
(1/2)mv_i^2 - [tex]\frac{GmM}{(r_i)}[/tex] = (1/2)mv_f^2 - [tex]\frac{GmM}{(r_f)}[/tex]
[tex]v_f = \sqrt{[(v_i)^2 + 2MG((1/r_f) - (1/r_i))]}[/tex]
Therefore
[tex]v_f = \sqrt{(2.1*10^4)^2 + 2(1.9891 * 10^{31})*(6.67 * 10^{-11})(\frac{1}{4.9} * 10^{10}) - (\frac{1}{2.5} * 10^{11}))} *20.408 *10^{12}[/tex]
[tex]v_f = \sqrt{(441000000) + 435.38 * 10^{7}}[/tex]
vf = 6.92 * 10^(4) m/s
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