Gravity on Venus and finding the sun's mass from orbital data

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1. The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth.

Compute the acceleration due to gravity on the surface of Venus from these data.
What is the weight of a 5.00-kg rock on the surface of Venus?

2. Venus orbits the sun in a nearly circular orbit. The radius of the orbit of Venus is 1.08x10^11 m, the period of Venus is 224.7 days(1.94 x 10^7 s)

Calculate the mass of the sun.

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** I know F_grav = m * g and F_grav = GM_Earth*m/d^2

M_Earth = 5.98 x 10^24 kg and R_Earth = 6.38 x 10^6 m

I conclude: M_Venus = 4.88x10^24 kg and R_Venus = 6.06 x 10^6 m

Please help. I'm stuck
 
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It's not necessary and probably not a good idea to calculate the mass and radius of Venus! Instead use ratios.

Basic gravity formula: [tex]F= \frac{GmM}{r^2}[/tex] where G is the "universal" gravitational constant, M the mass of the planet, m the mass of the falling object, and r is the radius of the planet. Since mg= F, [tex]g= \frac{GM}{r^2}[/tex].
In particular, for the Earth [tex]g_e= 9.81 m/s^2= \frac{GM_e}{r_e^2}[/tex] and for Venus [tex]g_v= \frac{GM_v}{r_v^2}[/tex].

Divide the second equation by the first and the "G" terms cancel:
[tex]\frac{g_v}{9.81}= \frac{M_v}{M_e}\left(\frac{r_e}{r_v}\right)^2[/tex]. You are given those ratios.
 
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but I did: g_v = [(M_v*r_v^2)/(M_e*r_e)]*9.81 = 4.60 x 10^7 m/s^2

what did I do wrong?
 
but I did [(M_v*M_e)/(r_v*r_e)^2]*9.81 = 7.22 m/s^2 what did I do wrong?
 
[(M_v/M_e)*(r_v*r_e)^2]*9.81 = 7.22 m/s^2

[(4.88x10^24 kg/5.98x10^24 kg)*(6.06x10^6 m*6.38x10^6 m)^2]*9.81 = 7.22 m/s^2
 
sorry it was [(4.88x10^24 kg/5.98x10^24 kg)*(6.06x10^6 m/6.38x10^6 m)^2]*9.81 = 7.22 m/s^2
 
that got me 0.816 * 0.902 * 9.81 = 7.22
 
MAPgirl23 said:
sorry it was [(4.88x10^24 kg/5.98x10^24 kg)*(6.06x10^6 m/6.38x10^6 m)^2]*9.81 = 7.22 m/s^2

The problem is in the bold part:
[(4.88x10^24 kg/5.98x10^24 kg)*(6.06x10^6 m/6.38x10^6 m)^2]*9.81 = 7.22 m/s^2

You are dividing Venus' radius by the Earth's radius, and you should be dividing the Earth's radius by Venus' radius.