How Strong Is the Sun's Gravity at Earth's Orbit Distance?

[guru]

What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help

 

[Tide]

The gravitational pull of the Sun on the Earth is the centripetal force keeping the Earth from flying off into the great beyond.

 

[geometer]

What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help

Find the mass of the sun, and the distance between the sun and the earth. Then, think about Newton's Law of Universal Gravitation.

 

[guru]

This is what i obtained
Sun's Mass: 1.99x10^30 kg,
earth-sun distance: 150x10^6 km.

law of gravitational = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)

is that right?

 

[pervect]

What is the acceleration due to gravity of the sun at the distance of the Earth's orbit?

i have no idea where to begin.
Help

Probably the simplest and most direct approach is to use Newton's law of gravity.

Do you know what that is?

 

[guru]

Probably the simplest and most direct approach is to use Newton's law of gravity.

Do you know what that is?

law of gravity = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)
right?

 

[Tide]

The acceleration due to the Sun's gravity at the location of the Earth is just

$$a = \frac {v^2}{R}$$

where v is the speed of the Earth around the Sun and R is the radius of Earth's orbit. You know R and you know the Earth travels a distance of $2\pi R$ in one year so you know its speed too.

 

[geometer]

law of gravity = Gm/r^2
G=6.67*10^(-11)

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)
right?

That should be right!