Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance of 2.5×1011

Question

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance of 2.5×1011 m from the center of the sun, what is its speed when at a distance of 4.9×1010 m .Express your answer using two significant figures.

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Verity 4 days 2021-07-22T19:32:47+00:00 1 Answers 2 views 0

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    2021-07-22T19:34:33+00:00

    Answer:

    v_f = 6.92 x 10^(4) m/s

    Explanation:

    From conservation of energy,

    E = (1/2)mv² – GmM/r

    Where M is mass of sun

    Thus,

    E_i = E_f will give;

    (1/2)mv_i² – GmM/(r_i) = (1/2)mv_f² – GmM/(r_f)

    m will cancel out to give ;

    (1/2)v_i² – GM/(r_i) = (1/2)v_f² – GM/(r_f)

    Let’s make v_f the subject;

    v_f = √[(v_i)² + 2MG((1/r_f) – (1/r_i))]

    G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²

    Mass of sun is 1.9891 x 10^(30) kg

    v_i = 2.1×10⁴ m/s

    r_i = 2.5 × 10^(11) m

    r_f = 4.9 × 10^(10) m

    Plugging in all these values, we have;

    v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) – (1/(2.5 × 10^(11)))] 20.408 e12

    v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]

    v_f = √[(441000000) + (435.38 x 10^(7))

    v_f = 6.92 x 10^(4) m/s

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