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A 1.5 kg object is located at a distance of 6.4 x 10^6 m from the center of a larger object whose mass is 6.0 x 10^24 kg what is the size of the force acting on the smaller object?

User Tatyana
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1 Answer

4 votes

Answer:

Approximately 2.4 x 10^-8 N.

Explanation:

The force acting on the smaller object can be calculated using the formula for gravitational force:

F = G * (m1 * m2) / d^2

Where F is the force, G is the gravitational constant (6.674 x 10^-11 N * m^2 / kg^2), m1 is the mass of the smaller object (1.5 kg), m2 is the mass of the larger object (6.0 x 10^24 kg), and d is the distance between the two objects (6.4 x 10^6 m).

Substituting these values into the formula, we get:

F = 6.674 x 10^-11 * (1.5 * 6.0 x 10^24) / (6.4 x 10^6)^2

We can simplify this expression by dividing both sides by 6.0 x 10^24 to get:

F / 6.0 x 10^24 = 6.674 x 10^-11 * (1.5 / (6.4 x 10^6)^2)

Then we can simplify the right-hand side by performing the calculations in parentheses:

F / 6.0 x 10^24 = 6.674 x 10^-11 * (1.5 / (6.4 x 10^6)^2)

= 6.674 x 10^-11 * (1.5 / 41.6 x 10^12)

= 6.674 x 10^-11 * 3.6 x 10^-13

Finally, we can multiply both sides by 6.0 x 10^24 to get the value of the force acting on the smaller object:

F = 6.0 x 10^24 * (6.674 x 10^-11 * 3.6 x 10^-13)

= 2.4 x 10^-8 N

Therefore, the size of the force acting on the smaller object is approximately 2.4 x 10^-8 N.

User Dylan Buth
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