When a falling meteoroid is at a distance above the Earth's surface of 3.00 times the Earth 's radius, what is the acceleration due to the Earth's gravitation?
Answer:
I suppose Re is the radius of Earth
a = MG/ (4 x Re)^2
= (9.8 m/s^2 )/16
= 0.613 m/s^2
I don't UNDERSTAND what MG becomes 9.8 ....
physics - Universal Gravitation 2?
2010-05-20 7:27 am
回答 (2)
2010-05-20 7:37 am
✔ 最佳答案
I don't UNDERSTAND why you replace why with what.It is actually M*G/Re^2 which becomes 9.8 m/s^2. That is the surface gravity of our planet (force per unit mass at its surface).
You can plug in actual values of M and G and the radius of Earth if you want to, and you should still get a similar answer. I got 0.6145 m/s^2 for the gravitational field at the asteroid's location, and for the Earth itself, my data would yield g=9.831 N/kg (about that experienced at the North pole).
2010-05-20 2:50 pm
g = GM / R^2
Here, R = (Re+3Re) = 4Re
So, g = GM / 16.Re^2
g = (9.8 / 16) = 0.6125 m/s^2
your work is correct
Here, R = (Re+3Re) = 4Re
So, g = GM / 16.Re^2
g = (9.8 / 16) = 0.6125 m/s^2
your work is correct
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