Two spacecraft in outer space attract each other with a force of 45 N. What would the attractive force be if they were \frac{1}{5} as far apart?
F_{attractive} = _____________N
(no rounding up)
if they were 1/5 as far apart?
25 times 45
same old 1/r^2
To solve this problem, we can use Newton's Law of Universal Gravitation, which states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The formula for the force of attraction is given by:
F = G * (m1 * m2) / r^2
Where:
F is the force of attraction
G is the gravitational constant (approximately 6.67430 x 10^-11 Nm^2/kg^2)
m1 and m2 are the masses of the objects
r is the distance between the objects
First, let's assume that the masses of the spacecrafts remain the same. So, the only variable that changes is the distance.
Let's say the original distance between the spacecrafts is represented by r1, and the new distance is represented by r2, where r2 = (1/5) * r1.
According to the question, the original force of attraction is 45 N.
Using the inverse square relationship, we can set up the following proportion:
45 N / r1^2 = F / r2^2
Now, substitute the value of r2:
45 N / r1^2 = F / [(1/5 * r1)^2]
Simplifying:
45 N / r1^2 = F / (1/25 * r1^2)
To get rid of the fractions, we can multiply both sides of the equation by r1^2:
45 N = F / (1/25)
Now, solve for F:
F = 45 N * (1/25)
F = 1.8 N
Therefore, the attractive force between the spacecrafts would be 1.8 N if they were 1/5th the original distance apart.