I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
The gravitational force would get stronger because the farther the two masses are separated the more gravitational force will be used to pull them together the closer they are the less gravitational pull is used to pull them together
Explanation:
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Answer:
W = 1,307 10⁶ J
Explanation:
Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)
F = G m₁ M / r²
W = ∫ F. dr
W = G m₁ M ∫ dr / r²
we integrate
W = G m₁ M (-1 / r)
We evaluate between the limits, lower r = R_ Moon and r = ∞
W = -G m₁ M (1 /∞ - 1 / R_moon)
W = G m1 M / r_moon
Body weight is
W = mg
m = W / g
The mass is constant, so we can find it with the initial data
For the capsule
m = 1000/32 = 165 / g_moon
g_moom = 165 32/1000
.g_moon = 5.28 ft / s²
I think it is easier to follow the exercise in SI system
W_capsule = 1000 pound (1 kg / 2.20 pounds)
W_capsule = 454 N
W = m_capsule g
m_capsule = W / g
m = 454 /9.8
m_capsule = 46,327 kg
Let's calculate
W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶
W = 1,307 10⁶ J
Answer:
Explanation:
Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².
Now, the pilot is traveling in a circle of radius
r = 3340 m
And the speed is
v = 495 m/s
Then, acceleration?
The acceleration of a circular motion can be determine using centripetal acceleration
a = v² / r
a = 495² / 3340
a = 73.36 m/s².
Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully
