The gravitational field strenght on the surface of a planet is 20N/kg; the planet has a radius of 1 x 10(4)km.
2006-08-29 00:06:23 · 7 answers · asked by jeanpace89 1 in Education & Reference ➔ Homework Help
Newton's Law of Universal Gravitation:
F = -G(m1m2)/r^2
where
F is the magnitude of the (repulsive) gravitational force between the two point masses
G is the gravitational constant (6.67 × 10^−11 N m2 kg−2)
m1 is the mass of the first point mass (5kg)
m2 is the mass of the second point mass (5.9736 * 10^24 kg)
r is the distance between the two point masses (Radius of Earth + height = 6.37101 * 10^6m + 1 * 10^7 m)
Note: You have to include the radius of earth, because gravity is held to originate from the center point.
Plug it in:
F = 6.67 × 10^−11 N m2 kg−2 * (5kg * 5.9736 * 10^24 kg) / (6.37101 * 10^6m + 1 * 10^7 m)^2
F = 6.67 × 10^−11 N m2 kg−2 * (2.9868 *10^25 kg^2) / (1.637101 * 10^7m)^2
F = 6.67 × 10^−11 N m2 kg−2 * (2.9868 *10^25 kg^2) / (2.6801 * 10^14 m^2)
F = 6.67 × 10^−11 N m2 kg−2 * 1.1144 * 10^11 kg^2/m^2
F = 6.67 × 10^−11 N m2 kg−2 * 1.1144 * 10^11 kg^2/m^2
F = 7.433048 N
2006-08-29 01:25:31 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
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2016-12-11 17:08:06 · answer #2 · answered by dricketts 4 · 0⤊ 0⤋
use newton's universal gravitational law ( Gravitational force = GMm/(r^2), M and m are the masses of which the gravitational force acts on each other. )
2006-08-29 01:19:42 · answer #3 · answered by vicks 2 · 0⤊ 0⤋
Why are you trying to be smart? Just ask a normal question will you. Are you an anorak???
2006-08-29 00:19:49 · answer #4 · answered by trog 3 · 0⤊ 1⤋
I know this, the answer is 42! lol
2006-08-29 00:31:08 · answer #5 · answered by Anonymous · 0⤊ 0⤋
This website is ideal for you:
http://www.antonine-education.co.uk/Physics_A2/Module_4/Topic_8/topic_8.htm
2006-08-29 00:27:09 · answer #6 · answered by Allasse 5 · 0⤊ 0⤋
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2006-08-29 00:10:47 · answer #7 · answered by Anonymous · 0⤊ 0⤋