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Math Help - Escape velocity

  1. #1
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    Escape velocity

    I am concerned on what is the velocity from the Sun at the distance of 1000 AU
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pfcbias View Post
    I am concerned on what is the velocity from the Sun at the distance of 1000 AU
    v_{esc}= \sqrt{\frac{GM_S}{r}}

    We have that r = 1000 AU and we need to convert that to m. So
    \frac{1000~AU}{1} \cdot \frac{149597870691~m}{1~AU} = 149597870691000~m \approx 1.496 \times 10^{14}~m

    And we know that
    M_S = 1.989 \times 10^{30}~kg

    Now plug in the numbers. For reference I get v_{esc} = 942~m/s

    -Dan
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  3. #3
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    Astronomy

    Thanks alot,
    Just 3 more questions....
    1. If circular speed at 1 AU from the Sun is 30 km/s. What is the circular speed at 20 AU?
    2. The acceleration of a 4 kg body to which an unbalanced 18 N force is applied
    3. The acceleration of gravity on Mars is 0.38 Earth gravities. How much would a dog that weighs 80 N on Earth weigh on Mars?
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by pfcbias View Post
    Thanks alot,
    Just 3 more questions....
    1. If circular speed at 1 AU from the Sun is 30 km/s. What is the circular speed at 20 AU?
    2. The acceleration of a 4 kg body to which an unbalanced 18 N force is applied
    3. The acceleration of gravity on Mars is 0.38 Earth gravities. How much would a dog that weighs 80 N on Earth weigh on Mars?
    Hints:
    1. Angular velocity doesn't change.
    2. F =m\cdot a
    3. G = m\cdot g
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by pfcbias View Post
    Thanks alot,
    Just 3 more questions....
    1. If circular speed at 1 AU from the Sun is 30 km/s. What is the circular speed at 20 AU?
    2. The acceleration of a 4 kg body to which an unbalanced 18 N force is applied
    3. The acceleration of gravity on Mars is 0.38 Earth gravities. How much would a dog that weighs 80 N on Earth weigh on Mars?
    Quote Originally Posted by wingless View Post
    Hints:
    1. Angular velocity doesn't change.
    2. F =m\cdot a
    3. G = m\cdot g
    The answer in red is wrong.

    Kepler's third law applies:

    \frac{\tau_1^2}{\tau_2^2}=\frac{r_1^3}{r_2^3}

    where \tau is the period and r the radius of an orbit.

    Translating this into terms of angular velocities gives:

    \frac{\omega_2^2}{\omega_1^2}=\frac{r_1^3}{r_2^3}

    where  \omega denotes an angular velocity


    RonL
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  6. #6
    Super Member wingless's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    The answer in red is wrong.

    Kepler's third law applies:

    \frac{\tau_1^2}{\tau_2^2}=\frac{r_1^3}{r_2^3}

    where \tau is the period and r the radius of an orbit.

    Translating this into terms of angular velocities gives:

    \frac{\omega_2^2}{\omega_1^2}=\frac{r_1^3}{r_2^3}

    where  \omega denotes an angular velocity


    RonL
    I thought your answer too. But the question doesn't state orbiting, which confused me. I may have misunderstood that.
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by wingless View Post
    I thought your answer too. But the question doesn't state orbiting, which confused me. I may have misunderstood that.
    Circluar speed at 1AU only has meaning for an orbit.

    RonL
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