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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
The gravitational pull of the Sun on the Earth is the centripetal force keeping the Earth from flying off into the great beyond.
geometer
Oct17-04, 10:06 PM
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.
This is what i obtained
Sun's Mass: 1.99x10^30 kg,
earth-sun distance: 150x10^6 km.
law of gravitional = Gm/r^2
G=6.67*10^(-11)
((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)
is that right?
pervect
Oct17-04, 11:31 PM
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?
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?
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
Oct18-04, 01:36 AM
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!
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