Newton's Law of Universal Gravitation can be used to solve for the force of gravity or the acceleration of gravity given enough information. You also must solve for one of the variables within the equation. Answer the following multiple choice problems to give yourself some more practice.
What is the magnitude of the gravitational force that acts on a 12 kg object and a 25 kg object that are 1.2 m apart?
- 1.67 x 10-8 N
- 1.39 x 10-8 N
- 250 N
- 208 N
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for force.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for force.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for force.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for force.
What is the magnitude of the gravitational force that acts on a bowling ball (7.2 kg, radius = 0.11 m) and a billiard ball (0.38 kg, radius = 0.028 m) if they are touching?
- 1.51 x 10-8 N
- 2.33 x 10-7 N
- 1.32 x 10-9 N
- 9.58 x 10-9 N
Remember to account for the distance between the centers of the balls, so you must add the radii and use that as r in \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\).
Remember to account for the distance between the centers of the balls, so you must add the radii and use that as r in \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\).
Remember to account for the distance between the centers of the balls, so you must add the radii and use that as r in \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\).
Remember to account for the distance between the centers of the balls, so you must add the raidii and use that as r in \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\).
The gravitational force between two masses is 100 N. If the magnitude of each mass doubles, what is the force between the masses?
- 25 N
- 50 N
- 200 N
- 400 N
If each mass doubles, the force will be four times greater.
If each mass doubles, the force will be four times greater.
If each mass doubles, the force will be four times greater.
If each mass doubles, the force will be four times greater.
The gravitational force between a black hole and a 58.2 kg person is 91500000 N. If they are 32500 m from each other find the mass of this black hole.
- 2.69 x 1023 kg
- 2.49 x 1025 kg
- 7.66 x 1020 kg
- 1.66 x 1015 kg
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for the missing mass.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for the missing mass.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for the missing mass.
Use \(\small\mathsf{ F_g = \frac{Gm_1m_2}{r^2} }\) to solve for the missing mass.
Find the acceleration due to gravity if you're at sea level. (mEarth = 5.98 x 1024 kg, rEarth = 6.37 x 106 m)
- 9.75 m/s2
- 9.80 m/s2
- 9.83 m/s2
- 9.92 m/s2
Use \(\small\mathsf{ g = \frac{GM}{r^2} }\) to find the acceleration of gravity.
Use \(\small\mathsf{ g = \frac{GM}{r^2} }\) to find the acceleration of gravity.
Use \(\small\mathsf{ g = \frac{GM}{r^2} }\) to find the acceleration of gravity.
Use \(\small\mathsf{ g = \frac{GM}{r^2} }\) to find the acceleration of gravity.
The gravitational force between two masses is 100 N. If the distance between the masses increases by a factor of 10, what is the force between the masses?
- 1 N
- 100 N
- 1000 N
- 10000 N
If the distance is increased by a factor of 10, the force would decrease by \(\small\mathsf{ \frac{1}{100} }\).
If the distance is increased by a factor of 10, the force would decrease by \(\small\mathsf{ \frac{1}{100} }\).
If the distance is increased by a factor of 10, the force would decrease by \(\small\mathsf{ \frac{1}{100} }\).
If the distance is increased by a factor of 10, the force would decrease by \(\small\mathsf{ \frac{1}{100} }\).
Summary
Questions answered correctly:
Questions answered incorrectly:
