Calculating Acceleration on the Surface of Venus

[SantiagoHill]

Friends, hello everyone!
Please help with this calculation.

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth.
(a) Compute the acceleration due to gravity on the surface of Venus from these data.
(b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venuse?

 

[topsquark]

Friends, hello everyone!
Please help with this calculation.

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth.
(a) Compute the acceleration due to gravity on the surface of Venus from these data.
(b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venuse?

a) You probably already have a formula for g' (the acceleration due to gravity) but here is the derivation anyway:
\(\displaystyle F = \dfrac{GMm}{R^2} = m g'\)

\(\displaystyle g' = \dfrac{GM}{R^2}\)

So if the mass of Venus is \(\displaystyle M = 0.815 M_E\) and \(\displaystyle R = 0.949 R_E\) then what is g'?

b) On Earth w = mg. So find m. On Venus w' = mg'...

-Dan

 

[WestleyColeman]

a) The mass of Venus MV represent 81.5% of the of the Earth. So, the mass of Venus is
MV=(81.5%)ME
(1)
Where ME is the mass of the Earth. Plug the mass of Earth into equation (1), so we can get MV by
MV=(0.815)ME=(0.815)(5.97×1024kg)=4.86×1024kg
The radius of Venus represent 94.9% of the radius of the Earth. So, the radius of Venus is
RV=(94.9%)RE
(2)
Where RE is the radius of the Earth. Plug the radius of Earth into equation (2), so we can get RV by
RV=(0.949)(6.38×106m)=6.05×106m
The weight w of a body is the total gravitational force exerted on it by all other bodies in the univerce where this gravitational force F is given by Newton`s general law of gravity by
F=GMVmRV2
(3)
Where RV is the radius of Venus, m is the mass of the body. As we mentioned above, this gravitational force equals the weight of the body which is mg, so using equation (3) we get the acceleration on Venus by
mg=GMVmRV2
gv=GMVRV2
(4)
Now, we plug values for RV, MV and G into equation (4) to get gv
gv=GMVRV2
=(6.67×10−11N⋅m2÷kg2)(4.86×1024kg)(6.05×106m)2

 

[topsquark]

a) The mass of Venus MV represent 81.5% of the of the Earth. So, the mass of Venus is
MV=(81.5%)ME
(1)
Where ME is the mass of the Earth. Plug the mass of Earth into equation (1), so we can get MV by
MV=(0.815)ME=(0.815)(5.97×1024kg)=4.86×1024kg
The radius of Venus represent 94.9% of the radius of the Earth. So, the radius of Venus is
RV=(94.9%)RE
(2)
Where RE is the radius of the Earth. Plug the radius of Earth into equation (2), so we can get RV by
RV=(0.949)(6.38×106m)=6.05×106m
The weight w of a body is the total gravitational force exerted on it by all other bodies in the univerce where this gravitational force F is given by Newton`s general law of gravity by
F=GMVmRV2
(3)
Where RV is the radius of Venus, m is the mass of the body. As we mentioned above, this gravitational force equals the weight of the body which is mg, so using equation (3) we get the acceleration on Venus by
mg=GMVmRV2
gv=GMVRV2
(4)
Now, we plug values for RV, MV and G into equation (4) to get gv
gv=GMVRV2
=(6.67×10−11N⋅m2÷kg2)(4.86×1024kg)(6.05×106m)2

You are missing some division operators but otherwise correct.

Here's a bit of a shortcut.

Define \(\displaystyle g' = \dfrac{GM_V}{R_V^2}\) and \(\displaystyle g = \dfrac{GM_E}{R_E^2}\). We know that \(\displaystyle M_V = 0.815 M_E\) and \(\displaystyle R_V = 0.949 R_E\). Then
\(\displaystyle g' = \dfrac{GM_V}{R_V^2} = \dfrac{G (0.815 M_E)}{(0.949 R_E)^2} = \dfrac{0.815}{0.949^2} \dfrac{G M_E}{R_E^2} = \dfrac{0.815}{0.949^2} \cdot g\)
and now you can use g = 9.81 m/s^2.

-Dan

 

[WestleyColeman]

You are missing some division operators but otherwise correct.

Here's a bit of a shortcut.

Define \(\displaystyle g' = \dfrac{GM_V}{R_V^2}\) and \(\displaystyle g = \dfrac{GM_E}{R_E^2}\). We know that \(\displaystyle M_V = 0.815 M_E\) and \(\displaystyle R_V = 0.949 R_E\). Then
\(\displaystyle g' = \dfrac{GM_V}{R_V^2} = \dfrac{G (0.815 M_E)}{(0.949 R_E)^2} = \dfrac{0.815}{0.949^2} \dfrac{G M_E}{R_E^2} = \dfrac{0.815}{0.949^2} \cdot g\)
and now you can use g = 9.81 m/s^2.

In order to get a better and more complete explanation of the solution of this problem, I advise you to go to https://plainmath.net/38641/the-mas...-the-earth-and-its-radius-is-94-9%-that-of-th and read the information here. In addition, you can find other useful information here.

Thanks for the advice.