Ticlieple The gravitational attraction force between two objects $$ m $$, and $$ 1 \mathrm{~m}_{2} $$, with distance of 2 m apart 4 N . How great will the gravitational attraction force between two objects change if: (1) the distance between them, change from 2 m to 4 m ? (2.) the distance between them change from 2 m to 1 m ? (3) both masses doubles, while the distance between then is halved? (4) both masses and the distance between them are doubled?

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To solve the problem, we will use Newton's law of universal gravitation, which states that the gravitational force $$ F $$ between two masses $$ m_1 $$ and $$ m_2 $$ separated by a distance $$ r $$ is given by the formula:

$$
F = G \frac{m_1 m_2}{r^2}
$$

where $$ G $$ is the gravitational constant.

Given that the gravitational force $$ F $$ is 4 N when the distance $$ r $$ is 2 m, we can analyze the changes in gravitational force under the specified conditions.

### Known Conditions
- Initial force $$ F_1 = 4 \, \text{N} $$
- Initial distance $$ r_1 = 2 \, \text{m} $$

### (1) Distance changes from 2 m to 4 m
When the distance is doubled, the new distance $$ r_2 = 4 \, \text{m} $$. The gravitational force can be calculated as follows:

$$
F_2 = G \frac{m_1 m_2}{(r_2)^2} = G \frac{m_1 m_2}{(4)^2} = G \frac{m_1 m_2}{16}
$$

Since $$ F_1 = G \frac{m_1 m_2}{(2)^2} = G \frac{m_1 m_2}{4} $$, we can relate $$ F_2 $$ to $$ F_1 $$:

$$
F_2 = \frac{F_1}{4} = \frac{4 \, \text{N}}{4} = 1 \, \text{N}
$$

### (2) Distance changes from 2 m to 1 m
When the distance is halved, the new distance $$ r_3 = 1 \, \text{m} $$. The gravitational force can be calculated as follows:

$$
F_3 = G \frac{m_1 m_2}{(r_3)^2} = G \frac{m_1 m_2}{(1)^2} = G \frac{m_1 m_2}{1}
$$

Using the relationship with $$ F_1 $$:

$$
F_3 = 4 \times F_1 = 4 \times 4 \, \text{N} = 16 \, \text{N}
$$

### (3) Both masses double, while the distance is halved
If both masses are doubled, the new masses are $$ 2m_1 $$ and $$ 2m_2 $$, and the distance is halved to $$ r_4 = 1 \, \text{m} $$:

$$
F_4 = G \frac{(2m_1)(2m_2)}{(1)^2} = 4G \frac{m_1 m_2}{1} = 4F_1 = 4 \times 4 \, \text{N} = 16 \, \text{N}
$$

### (4) Both masses and the distance are doubled
If both masses are doubled and the distance is also doubled, the new masses are $$ 2m_1 $$ and $$ 2m_2 $$, and the new distance is $$ r_5 = 4 \, \text{m} $$:

$$
F_5 = G \frac{(2m_1)(2m_2)}{(4)^2} = G \frac{4m_1 m_2}{16} = \frac{G \frac{m_1 m_2}{4}}{4} = \frac{F_1}{4} = \frac{4 \, \text{N}}{4} = 1 \, \text{N}
$$

### Summary of Results
1. Distance changes from 2 m to 4 m: $$ F_2 = 1 \, \text{N} $$
2. Distance changes from 2 m to 1 m: $$ F_3 = 16 \, \text{N} $$
3. Both masses double, distance halved: $$ F_4 = 16 \, \text{N} $$
4. Both masses and distance doubled: $$ F_5 = 1 \, \text{N} $$
When the distance between two objects increases from 2 meters to 4 meters, the gravitational force decreases to 1 N. If the distance decreases from 2 meters to 1 meter, the force increases to 16 N. Doubling both masses while halving the distance also results in a force of 16 N. Doubling both masses and the distance brings the force back to 1 N.

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