# Comparing escape velocities on Earth and Mars

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

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

Fibonacci

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

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

Mars

so u mean im right? check the equation if it is right??

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.

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

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.

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.

whoozum

but the correct answer in my book is 0.488

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.

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

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

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?

ull just have to subtract the distance to the answer.... but i dont know the distance

whozum, i still dont get it.. wat will i do??

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You can look up the radius of the earth

It is 6.3 x 10^6 m

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

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

VietDao29
Homework Helper
I dunno how come you use the equation to find the escape velocity here:
$$v_{esc} = \sqrt{\frac{2GM}{R}}$$
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)
Hope it help.
Viet Dao,

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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
Homework Helper
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