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Comparing escape velocities on Earth and Mars

  1. Apr 2, 2005 #1
    The problem is..
    the mass of MArs is 0.107 times Earth's mass and its raduis is 0.532 times Earth's raduis. how does the escape velocity on Mars compare to the velocity on Earth?

    i did this..
    Vof escape = square root of 2G M(mars)/r(Mars)
    Vof escape is proportional to the square root of M(mars)/r(Mars)
    Vof Mars/Vof Earth = square root of 0.107x M(earth)/0.532x r(earth)
    square root of M(earth)/r(earth)
    Vof Mars/Vof Earth = square root of 0.107x M(earth)/0.532x r(earth) *square root of r(earth)/M(earth)
    Vof Mars = 0.448*Vof Earth

    but the answer in my book is .488 x V of Eath... im not sure if im right or the book is right
     
    Last edited: Apr 2, 2005
  2. jcsd
  3. Apr 2, 2005 #2
    What? Was that a question? I can read minds, but not over the internet.

    Fibonacci
     
  4. Apr 2, 2005 #3
    The problem is..
    the mass of MArs is 0.107 times Earth's mass and its raduis is 0.532 times Earth's raduis. how does the escape velocity on Mars compare to the velocity on Earth?

    i did this..
    Vof escape = square root of 2G M(mars)/r(Mars)
    Vof escape is proportional to the square root of M(mars)/r(Mars)
    Vof Mars/Vof Earth = square root of 0.107x M(earth)/0.532x r(earth)
    square root of M(earth)/r(earth)
    Vof Mars/Vof Earth = square root of 0.107x M(earth)/0.532x r(earth) *square root of r(earth)/M(earth)
    Vof Mars = 0.448*Vof Earth

    but the answer in my book is .488 x V of Eath... im not sure if im right or the book is right
     
  5. Apr 2, 2005 #4
    Your answer does indeed seem to be correct. The answer in the book is most likely a typo.
     
  6. Apr 2, 2005 #5
    Mars

    so u mean im right? check the equation if it is right??
     
  7. Apr 2, 2005 #6
    [tex]\frac{v_m}{v_e}={\sqrt{\frac{M_m R_e}{M_e R_m}}={\sqrt{\frac{0,107 M_e R_e}{0,532 M_e R_e}}[/tex]

    This does indeed equal 0,448.
     
    Last edited: Apr 2, 2005
  8. Apr 2, 2005 #7
    [tex] v_{esc} = \sqrt{\frac{2GM}{R}} [/tex]

    Me = 5.98 x 10^24
    Re = 6.3 x 10^6m
    Mm = 6.4 x 10^23kg
    Rm = 3.4 x 10^6m

    Earth: [tex] v_{esc} = \sqrt{\frac{2(6.67 x 10^{-11})(5.98 x 10^{24})}{6.3 x 10^6}} = 11252.73m/s [/tex]

    Mars: [tex] v_{esc} = \sqrt{\frac{2(6.67 x 10^{-11})(6.4 x 10^{23})}{3.4 x 10^6}} = 5011.047m/s [/tex]

    [tex] \frac{v_{mars}}{v_{earth}} = \frac{5011.047}{11252.73} = 0.446[/tex]

    Its a matter of significant figures.
     
  9. Apr 2, 2005 #8
    thanks

    ok... thanks a lot for the big help...

    another problem is... a rocket is launched vertically from earth's surface with a velocity of3.4km/s. How high it go a)from Earth's centre and b) from Earth's surface.
    a)r = 2GM
    V^2
    = 2(6.673x10^-11N m^2/kg^2) (5.98 x10^24kg)
    (3400m/s)^2
    r=7.0x107m

    i dont know what to do with b), i am thinking to add the raduis of the earth, but my answer is wrong... the answer should be 650km.
     
  10. Apr 2, 2005 #9
    whoozum

    but the correct answer in my book is 0.488
     
  11. Apr 2, 2005 #10
    ok guys...

    another problem is... a rocket is launched vertically from earth's surface with a velocity of3.4km/s. How high it go a)from Earth's centre and b) from Earth's surface.
    a)r = 2GM
    V^2
    = 2(6.673x10^-11N m^2/kg^2) (5.98 x10^24kg)
    (3400m/s)^2
    r=7.0x107m

    i dont know what to do with b), i am thinking to add the raduis of the earth, but my answer is wrong... the answer should be 650km.
     
  12. Apr 2, 2005 #11
    If its 7 x 10^7 from earth's center, thenthe rocket is closer to the surface than the center.
     
  13. Apr 2, 2005 #12
    so what will i do then if it is closer to the earth's surface, the information is so limited?
     
  14. Apr 2, 2005 #13
    so, will i add 7x10^7 to the raduis of the earth?
     
  15. Apr 2, 2005 #14
    Center ----------------- Surface ---------------------------Rocket

    If you know the distance from the rocet to the center, and from the surface to the center. How do you find the distance from the rocket to the sufrace?
     
  16. Apr 2, 2005 #15
    ull just have to subtract the distance to the answer.... but i dont know the distance
     
  17. Apr 2, 2005 #16
    whozum, i still dont get it.. wat will i do??
     
    Last edited: Apr 2, 2005
  18. Apr 2, 2005 #17
    You can look up the radius of the earth

    It is 6.3 x 10^6 m
     
  19. Apr 2, 2005 #18
    rocket

    i know it is... will i do this?

    change of raduis = r of rocket from earth - r of earth
    7x10^7 - 6.38x10^6
    63620000m

    my answer is 63620000m... but in the book, the answer is 650km
     
    Last edited: Apr 2, 2005
  20. Apr 2, 2005 #19
    I don't think were using the right equation, either that or the answer in the book is incorrect.
     
  21. Apr 3, 2005 #20

    VietDao29

    User Avatar
    Homework Helper

    I dunno how come you use the equation to find the escape velocity here:
    [tex]v_{esc} = \sqrt{\frac{2GM}{R}}[/tex]
    Try using integral instead:
    Draw a graph of:
    [tex]F_{gra} = \frac{GmM}{r^{2}}[/tex]
    The integration of F with respect to r is the work done by the Earth.
    So
    [tex]\frac{mv^{2}}{2} = -W_{Earth} = \int{\frac{GmM}{r^{2}}dr}[/tex]
    With x-axis is r and y-axis is F
    Now you have
    [tex]\frac{v^{2}}{2} = \int^{H}_{r_{0}}{\frac{GM}{r^{2}}dr}[/tex]
    [tex]=-\frac{GM}{H} + \frac{GM}{r_{0}}[/tex]
    H is the height from the earth center the setellite will go.
    M is Earth's mass (6. 10^24 kg)
    m is satellite's mass.
    ro is the radius of the Earth (6400 km = 6400000 m)
    v is the satellite's initial valocity (3400m/s)
    The answer will be about 650 km.
    Hope it help.
    Viet Dao,
     
    Last edited: Apr 3, 2005
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