A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit of Mercury), at which point its speed is 9.5 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.)

Answers

Answer 1
Answer:

Answer:

58515.9 m/s

Explanation:

We are given that

d_1=4.7* 10^(10) m

v_i=9.5* 10^4 m/s

d_2=6* 10^(12) m

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

K_i+U_i=K_f+U_f

(1)/(2)mv^2_i-(GmM)/(d_1)=(1)/(2)mv^2_f-(GmM)/(d_2)

(1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2)=(1)/(2)v^2_f

v^2_f=2((1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2))

v_f=\sqrt{2((1)/(2)v^2_i-(GM)/(d_1)+(GM)/(d_2))}

Using the formula

v_f=\sqrt{v^2_i+2GM((1)/(d_2)-(1)/(d_1))}

v_f=\sqrt{(9.5* 10^4)^2+2* 6.7* 10^(-11)* 1.98* 10^(30)((1)/(6* 10^(12))-(1)/(4.7* 10^(10)))

Where mass of sun=M=1.98* 10^(30) kg

G=6.7* 10^(-11)

v_f=58515.9 m/s


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Answers

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Answers

Answer:

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Explanation:

It is given that,

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Answers

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Answers

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Answers

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