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Qmechanic ♦
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corrected spelling , fixed grammar , improved formatting, added tags , Edited with LATEX
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UKH
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Understanding Need help in understanding accelaration due to gravity

I've been having some trouble in understanding acceleration due to gravity.

On earth, the acceleration due to gravity is an average of 9.80$\frac{m}{s^2}$$9.80$ $m/s^2$. The mass of the earth is approximately 5.972 × 1024 kg$5.972 \times 10^{24} kg$. The acceleration due to gravity on the surface of the sun is 273.7$\frac{m}{s^2}$,$273.7$ $m/s^2$ and its Massmass is about 1.989 × 1030$1.989 \times 10^{30}$.

So: $$ \frac{Mass of Sun}{Accelaration at Sun'surface} = \frac{Mass of Earth}{Accelaration at Earth'surface}$$

$$ \frac{\textrm{Mass of Sun}}{\textrm{Accelaration at Sun surface}} = \frac{\textrm{Mass of Earth}}{\textrm{Accelaration at Earth surface}}$$

Why don't the above numbers equal each other? Is it because I am doing mass divided by acceleration?

Understanding accelaration due to gravity

I've been having some trouble in understanding acceleration due to gravity.

On earth, the acceleration due to gravity is an average of 9.80$\frac{m}{s^2}$. The mass of the earth is approximately 5.972 × 1024 kg. The acceleration due to gravity on the surface of the sun is 273.7$\frac{m}{s^2}$, and its Mass is 1.989 × 1030.

So: $$ \frac{Mass of Sun}{Accelaration at Sun'surface} = \frac{Mass of Earth}{Accelaration at Earth'surface}$$

Why don't the above numbers equal each other? Is it because I am doing mass divided by acceleration?

Need help in understanding accelaration due to gravity

I've been having some trouble in understanding acceleration due to gravity.

On earth, the acceleration due to gravity is an average of $9.80$ $m/s^2$. The mass of the earth is approximately $5.972 \times 10^{24} kg$. The acceleration due to gravity on the surface of the sun is $273.7$ $m/s^2$ and its mass is about $1.989 \times 10^{30}$.

So

$$ \frac{\textrm{Mass of Sun}}{\textrm{Accelaration at Sun surface}} = \frac{\textrm{Mass of Earth}}{\textrm{Accelaration at Earth surface}}$$

Why don't the above numbers equal each other? Is it because I am doing mass divided by acceleration?

corrected spelling , fixed grammar , improved formatting, added tags , Edited with LATEX
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edit approved Jan 1, 2017 at 11:24
InquisitiveMind
edit approved Jan 1, 2017 at 11:24
InquisitiveMind
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I need help understanding Understanding accelaration due to gravity

I've been having some trouble in understanding acceleration due to gravity. 

On earth, the acceleration due to gravity is an average of 9.80m/s^29.80$\frac{m}{s^2}$. The mass of the earth is approximately 5.972 × 10^24 kg5.972 × 1024 kg. The acceleration due to gravity on the surface of the sun is 273.7m/s^2273.7$\frac{m}{s^2}$, and its massMass is 1.989 × 10^30 kg1.989 × 1030. So

So: m/a=m/a 5.972 x 10^24 kg/9.80m/s^2 = 1.989 x 10^30 kg/273.7m/s^2 6.09=/=7.267 Why$$ \frac{Mass of Sun}{Accelaration at Sun'surface} = \frac{Mass of Earth}{Accelaration at Earth'surface}$$

Why don't the above numbers equal each other? Is it because iI am doing mass divided by acceleration?

I need help understanding gravity

I've been having some trouble understanding acceleration due to gravity. On earth, the acceleration due to gravity is an average of 9.80m/s^2. The mass of earth is approximately 5.972 × 10^24 kg. The acceleration due to gravity on the surface of the sun is 273.7m/s^2, and its mass is 1.989 × 10^30 kg. So: m/a=m/a 5.972 x 10^24 kg/9.80m/s^2 = 1.989 x 10^30 kg/273.7m/s^2 6.09=/=7.267 Why don't the above numbers equal each other? Is it because i am doing mass divided by acceleration?

Understanding accelaration due to gravity

I've been having some trouble in understanding acceleration due to gravity. 

On earth, the acceleration due to gravity is an average of 9.80$\frac{m}{s^2}$. The mass of the earth is approximately 5.972 × 1024 kg. The acceleration due to gravity on the surface of the sun is 273.7$\frac{m}{s^2}$, and its Mass is 1.989 × 1030.

So: $$ \frac{Mass of Sun}{Accelaration at Sun'surface} = \frac{Mass of Earth}{Accelaration at Earth'surface}$$

Why don't the above numbers equal each other? Is it because I am doing mass divided by acceleration?

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Alex Ambruso
Alex Ambruso
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