no idea.
where's your work?
Three spheres are arranged in the xy plane as in the figure below. The first sphere, of mass
m1 = 13.4 kg,
is located at the origin; the second sphere, of mass
m2 = 4.56 kg
is located at (−6.00, 0.00) m; and the third sphere, of mass
m3 = 4.00 kg
is located at (0.00, 5.00) m. Assuming an isolated system, what is the net gravitational force on the sphere located at the origin?
I keep getting 1.82502801*10^(-10). Am I wrong?
3 answers
6.67 x 10⁻¹¹(13.4*4.56)/-6^2 = 1.13280026*10^(-10)
6.67 x 10⁻¹¹(13.4*4)/5^2 = 1.4309056*10^(-10)
\sqrt((1.4309056*10^(-10))^(2)+(-1.13280026*10^(-10))^(2)) = 1.82502801*10^(-10).
6.67 x 10⁻¹¹(13.4*4)/5^2 = 1.4309056*10^(-10)
\sqrt((1.4309056*10^(-10))^(2)+(-1.13280026*10^(-10))^(2)) = 1.82502801*10^(-10).
Fx = - G m2 m1 / 36 = - G m1 (m2/36)
Fy = + G m3 m1 / 25 = G m1 (m3/25)
|F| = sqrt (Fx^2 + Fy^2)
G = 6.67*10^-11
G m1 = 6.67*10^-11 * 13.4 = 8.94 *10^-10
Fx= -G m1 (m2/36) = - 8.94*10^-10 * 0.127 = - 1.14*10^-10
Fy =+G m1(m3/25) = +8.94 *10^-10 * 0.16 = + 1.43 * 10^-10
so 10^-10 sqrt (1.14^2 + 1.43^2)\\\
10^-10 sqrt (1.3 + 2.05) = 1.83 * 10^-10
I agree with you.
Fy = + G m3 m1 / 25 = G m1 (m3/25)
|F| = sqrt (Fx^2 + Fy^2)
G = 6.67*10^-11
G m1 = 6.67*10^-11 * 13.4 = 8.94 *10^-10
Fx= -G m1 (m2/36) = - 8.94*10^-10 * 0.127 = - 1.14*10^-10
Fy =+G m1(m3/25) = +8.94 *10^-10 * 0.16 = + 1.43 * 10^-10
so 10^-10 sqrt (1.14^2 + 1.43^2)\\\
10^-10 sqrt (1.3 + 2.05) = 1.83 * 10^-10
I agree with you.