net gravitational foce on equilatorial triangle on every point
Answers
Answered by
1
Answer:
F
net
=
(F
1
2
+F
2
2
+2F
1
F
2
cosθ)
Here, F
1
=F
2
=
r
2
Gmm
and θ=60
0
Therefore, on solving we get, F
net
=
3
a
2
Gm
2
Answered by
3
Answer:
ANSWER
F
net
=
(F
1
2
+F
2
2
+2F
1
F
2
cosθ)
Here, F
1
=F
2
=
r
2
Gmm
and θ=60
0
Therefore, on solving we get, F
net
=
3
a
2
Gm
2
Explanation:
hope it helps u....
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