Circular Motion and Gravitation Practice Questions
AP (Advanced Placement) · AP Physics 1 · 151 free MCQs with instant results and detailed explanations.
151
Total
55
Easy
71
Medium
25
Hard
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Sample Questions from Circular Motion and Gravitation
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Q1
Easy
What is the centripetal acceleration of an object moving in a circle of radius 5 m at a constant speed of 20 m/s?
A.
80 m/s²
B.
40 m/s²
C.
20 m/s²
D.
10 m/s²
Show Answer & Explanation
Correct Answer: A
Centripetal acceleration (a_c) is given by the formula a_c = v²/r. Here, v = 20 m/s and r = 5 m. Thus, a_c = (20 m/s)² / (5 m) = 400 m²/s² / 5 m = 80 m/s².
Q2
Easy
A satellite is in a circular orbit around Earth at a height where the gravitational acceleration is 9.8 m/s². What is the ratio of the satellite's orbital speed to that of an object falling freely from the same height?
A.
1:1
B.
√2:1
C.
2:1
D.
3:1
Show Answer & Explanation
Correct Answer: A
In a circular orbit, the orbital speed (v) of the satellite is equal to the speed an object would have when falling freely under gravity (g), hence the ratio is 1:1.
Q3
Easy
If the radius of a circular path is doubled, how does the centripetal force required for a constant speed change?
A.
It doubles.
B.
It remains the same.
C.
It is halved.
D.
It quadruples.
Show Answer & Explanation
Correct Answer: A
Centripetal force (F_c) is given by F_c = mv²/r. If the radius (r) is doubled, to keep the same speed (v), the force must double because F_c is inversely proportional to r.
Q4
Medium
A satellite orbits Earth in a circular path with a radius of 8000 km. Calculate the gravitational force acting on a 500 kg satellite. (Use G = 6.67 x 10^-11 N m²/kg², mass of Earth = 5.97 x 10^24 kg)
A.
3.14 N
B.
4.90 N
C.
6.67 N
D.
9.81 N
Show Answer & Explanation
Correct Answer: B
Using the formula F = G(m1*m2)/r², we can plug in the values: F = (6.67 x 10^-11)(5.97 x 10^24)(500) / (8000 x 10^3)² = 4.90 N. Thus, the gravitational force is 4.90 N.
Q5
Medium
An object is moving in a circular path of radius 2 m with a constant speed of 4 m/s. What is the centripetal acceleration of the object?
A.
2 m/s²
B.
4 m/s²
C.
8 m/s²
D.
16 m/s²
Show Answer & Explanation
Correct Answer: B
The formula for centripetal acceleration is a_c = v²/r. Substituting v = 4 m/s and r = 2 m gives a_c = (4)² / 2 = 8 / 2 = 4 m/s².
Q6
Medium
A satellite is orbiting Earth in a circular path with a radius of 7000 km. What is the gravitational force acting on a 1000 kg satellite? (Use G = 6.67 × 10^-11 N m²/kg² and mass of Earth = 5.97 × 10^24 kg)
A.
9.14 N
B.
5.67 N
C.
7.12 N
D.
2.67 N
Show Answer & Explanation
Correct Answer: A
The gravitational force can be calculated using F = G(m1*m2)/r². Plugging in the values gives approximately 9.14 N, confirming option A as correct.
Q7
Medium
Two planets are in a circular orbit around a star. Planet A has a mass of 3 × 10^24 kg, and Planet B has a mass of 6 × 10^24 kg. If both planets orbit at the same radius R, what is the ratio of their gravitational forces exerted by the star on each planet?
A.
1:2
B.
2:1
C.
1:1
D.
4:1
Show Answer & Explanation
Correct Answer: A
The gravitational force depends on the mass of the planets. Since the force for each planet is proportional to its mass and both have different masses, the ratio is 1:2, aligning with option A.
Q8
Hard
Two objects with masses m1 = 2 kg and m2 = 3 kg are separated by a distance of 1 meter. Calculate the gravitational force between them. (Use G = 6.67 × 10^-11 N m²/kg²)
A.
4.00 × 10^-11 N
B.
6.67 × 10^-11 N
C.
2.00 × 10^-11 N
D.
8.01 × 10^-11 N
Show Answer & Explanation
Correct Answer: A
The gravitational force can be calculated using Newton's law of gravitation: F = G * (m1 * m2) / r². Here, F = (6.67 × 10^-11 N m²/kg²) * (2 kg * 3 kg) / (1 m)² = 4.00 × 10^-11 N.
Q9
Hard
Two objects, A and B, are in a circular motion where A has a mass of 2 kg and is orbiting with a radius of 5 m at a speed of 10 m/s, while B has a mass of 4 kg and is orbiting with a radius of 10 m at a speed of 5 m/s. Which object experiences greater centripetal acceleration?
A.
Object A
B.
Object B
C.
Both experience the same
D.
Cannot be determined
Show Answer & Explanation
Correct Answer: A
Centripetal acceleration is given by a_c = v²/r. For object A, a_cA = (10 m/s)² / 5 m = 20 m/s². For object B, a_cB = (5 m/s)² / 10 m = 2.5 m/s². Thus, object A experiences greater centripetal acceleration.
Q10
Hard
A satellite is in a circular orbit around the Earth at an altitude where the gravitational field strength is one-fourth of that at the Earth's surface. What is the ratio of the orbital speed of the satellite to the speed of an object in free fall at the Earth's surface?
A.
1
B.
2
C.
4
D.
0.5
Show Answer & Explanation
Correct Answer: B
The orbital speed of the satellite is greater than the speed of free fall on the surface due to the reduced gravitational field strength. The orbital speed can be found using the formula v = √(g_r * r), where g_r is the gravitational field at orbit and r is the radius from the center of the Earth. Since the gravitational field is one-fourth, the orbital speed is √(g/4) = (1/2)√g. This means the satellite's speed is √2 times faster than free fall, resulting in a ratio of 2.
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Circular Motion and Gravitation — AP (Advanced Placement) AP Physics 1 Practice Questions Online
This page contains 151 practice MCQs for the chapter Circular Motion and Gravitation in AP (Advanced Placement) AP Physics 1. The questions are organized by difficulty — 55 easy, 71 medium, 25 hard — so you can choose the right level for your preparation.
Every question includes a detailed explanation to help you understand the concept, not just memorize answers. Take a timed quiz to simulate exam conditions, or practice at your own pace with no time limit.