Respuesta :
Find F = Gm1m2/r²
where
G= 6.67×10^-11 N • m2/kg2
m1= 5.0×10^16 kg,
m2= 2.8×10^6 kg
and
r = ((20*10^3)²+(30*10^3)²)^(1/2)
Then,
F= 7183.07692 N
Fnet = (2)*F*cos[arctan(20/30)] = 11953 N
where
G= 6.67×10^-11 N • m2/kg2
m1= 5.0×10^16 kg,
m2= 2.8×10^6 kg
and
r = ((20*10^3)²+(30*10^3)²)^(1/2)
Then,
F= 7183.07692 N
Fnet = (2)*F*cos[arctan(20/30)] = 11953 N
Answer:
The net gravitational force exerted by the asteroids on the spaceship when the spaceship is 30 km away from that midpoint is 12000 Newtons
Explanation:
Gravitational force is given by
[tex]F=\frac{Gm_1m_2}{r^{2}}[/tex]
where [tex]m_1=2.8\times 10^{6} kg,m_2=5.0\times 10^{16}kg, r=[20^{2}+30^{2}]^{\frac{1}{2}}km=36km[/tex]
=>[tex]F=\frac{6.674\times 10^{-11}\times 2.8\times 10^{6}\times 5.0\times 10^{16}}{36000^{2}}N=7210N[/tex]
Now the force F is acting on spaceship due to two asteroids at an angle [tex]\Theta[/tex] with the line joining spaceship and mid-point such that [tex]\cos (\Theta )=\frac{30}{36}[/tex]
Therefore net force , [tex]F_net=2F\cos (\Theta )=2\times 7210\times \frac{30}{36}N=12017 N\sim 12000 N[/tex]
Thus the net gravitational force exerted by the asteroids on the spaceship when the spaceship is 30 km away from that midpoint is 12000 Newtons