Comparing escape velocities on Earth and Mars

[six789]

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... I am not sure if I am right or the book is right

 

[1123581321]

What? Was that a question? I can read minds, but not over the internet.

Fibonacci

 

[six789]

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... I am not sure if I am right or the book is right

 

[Berislav]

Your answer does indeed seem to be correct. The answer in the book is most likely a typo.

 

[six789]

Mars

so u mean I am right? check the equation if it is right??

 

[Berislav]

check the equation if it is right??

\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}}

This does indeed equal 0,448.

 

[whozum]

v_{esc} = \sqrt{\frac{2GM}{R}}

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

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

\frac{v_{mars}}{v_{earth}} = \frac{5011.047}{11252.73} = 0.446

Its a matter of significant figures.

 

[six789]

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 don't 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.

 

[six789]

whoozum

but the correct answer in my book is 0.488

 

[six789]

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 don't 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.

 

[whozum]

If its 7 x 10^7 from Earth's center, thenthe rocket is closer to the surface than the center.

 

[six789]

so what will i do then if it is closer to the Earth's surface, the information is so limited?

 

[six789]

so, will i add 7x10^7 to the raduis of the earth?

 

[whozum]

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?

 

[six789]

ull just have to subtract the distance to the answer... but i don't know the distance

 

[six789]

whozum, i still don't get it.. wat will i do??

 

[whozum]

You can look up the radius of the earth

It is 6.3 x 10^6 m

 

[six789]

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

 

[whozum]

I don't think were using the right equation, either that or the answer in the book is incorrect.

 

[VietDao29]

I don't know how come you use the equation to find the escape velocity here:
v_{esc} = \sqrt{\frac{2GM}{R}}
Try using integral instead:
Draw a graph of:
F_{gra} = \frac{GmM}{r^{2}}
The integration of F with respect to r is the work done by the Earth.
So
\frac{mv^{2}}{2} = -W_{Earth} = \int{\frac{GmM}{r^{2}}dr}
With x-axis is r and y-axis is F
Now you have
\frac{v^{2}}{2} = \int^{H}_{r_{0}}{\frac{GM}{r^{2}}dr}
=-\frac{GM}{H} + \frac{GM}{r_{0}}
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,

 

[whozum]

If you didn't notice we just skipped all the steps behind that.

v^2/2 = GM/r

Solve for r = (2GM/v^2)^.5

Doing that for Earth's radius and subtracting that from distance to center gives the answer we want.

 

[VietDao29]

If you do that, then... you will find the radius of some planet, which has the same mass as Earth, and has the escape velocity of 3,4 km/s. So obviously, that's not the correct equation!
If you do likle me, you will have a completely different answer (i.e: 650 km)
Viet Dao,

 

[tony873004]

I'd check the formula you're using. That formula doesn't take into account that your starting point is the surface. 6371 km needs to be in there somewhere.

You also need a formula that takes into account the diminishing strength of Earth gravity, or kinmatics would be all you'd need.

 

[six789]

heya tony, it is too complicated wat u are saying, we are not there yet in integral stuffs..

 

[tony873004]

Sorry, I think I posted that in the wrong thread. I was trying to answer the rocket at 3.4 km/s question.

 

[six789]

yeah it is.. the problem about...
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.

...that is the question that i make.

 

[whozum]

Using \frac{v^2}{2} =-\frac{GM}{H} + \frac{GM}{r}

Solving for h: \frac{2GMr}{v^2r-2GM}

G = 6.67e-11
M = 5.98e24kg
v=3.4e3m/s
r=6.4e6m

h = \frac{2(6.67x10^{-11})(5.98x10^{24})(6.4x10^6)}{(3.4x10^3)^2(6.4x10^6)-2(6.67x10^{-11})(5.98x10^{24})} = 7054229.925 = 7.05 x 10^6m from Earth's center.

h-r is the distance from the surface which is 650000m

 

[six789]

whozum, thanks for the help... =)

 

[whozum]

Sorry for the confusion

 

[six789]

u did a good job... heheh