physics : class11 Variation of Force with Altitude : Gravitational Force : Gravitation MCQ Questions and Answers
Practice important
Variation of Force with Altitude
MCQ questions and answers.
Page 1
of
1
|
10
MCQ Questions
1. A body of fixed mass is moved from the Earth's surface to an altitude equal to the Earth's radius R. What fraction of its weight at the surface does it have at this altitude?
A.
1/2
B.
1/4
C.
1/8
D.
1/9
Correct Answer:
1/4
Explanation:
The gravitational force at altitude h is F(h) = F0[R/(R+h)]². For h = R, F/F0 = [R/(2R)]² = 1/4.
2. At what altitude above the Earth's surface will the gravitational force on a body become 64% of its surface value? Let R be the Earth's radius.
A.
R/5
B.
R/4
C.
R/3
D.
2R/5
Correct Answer:
R/4
Explanation:
Using F/F0 = [R/(R+h)]², we get 0.64 = [R/(R+h)]². Thus R/(R+h) = 0.8, giving R+h = 1.25R and h = R/4.
3. A satellite is at an altitude R/2 above the Earth's surface, where R is the Earth's radius. The gravitational force on a body there compared with its surface value is:
A.
2/9
B.
1/3
C.
4/9
D.
1/2
Correct Answer:
4/9
Explanation:
For h = R/2, F/F0 = [R/(R+R/2)]² = [R/(3R/2)]² = (2/3)² = 4/9.
4. At an altitude h = R/3, the gravitational force on a body is what fraction of its surface value? What is the force at altitude 2R/3 as a fraction of the surface value?
A.
9/16 and 9/25
B.
3/4 and 3/5
C.
1/2 and 1/4
D.
4/9 and 9/16
Correct Answer:
9/16 and 9/25
Explanation:
For h = R/3, F/F0 = [R/(4R/3)]² = 9/16. For h = 2R/3, F/F0 = [R/(5R/3)]² = 9/25.
5. At what altitude above the Earth's surface is the gravitational force reduced by 19% from its surface value? Let R denote the Earth's radius.
A.
R/10
B.
R/9
C.
R/8
D.
R/6
Correct Answer:
R/9
Explanation:
A 19% reduction means F/F0 = 0.81. Therefore R/(R+h) = 0.9, so R+h = R/0.9 = 10R/9. Hence h = R/9.
6. The gravitational force on a body at altitude R above the Earth's surface is how many times the force on the same body at altitude 3R?
A.
2
B.
3
C.
4
D.
8
Correct Answer:
4
Explanation:
At h = R, F1/F0 = 1/4. At h = 3R, F2/F0 = [R/(4R)]² = 1/16. Thus F1/F2 = (1/4)/(1/16) = 4.
7. A body is first raised to an altitude x and then to an altitude 2x above the Earth's surface. If the gravitational force at x is 25/16 times the force at 2x, what is x in terms of the Earth's radius R?
A.
R/4
B.
R/3
C.
R/2
D.
2R/3
Correct Answer:
R/3
Explanation:
The force ratio is F(x)/F(2x) = [(R+2x)/(R+x)]² = 25/16. Hence (R+2x)/(R+x) = 5/4. Solving gives 4R + 8x = 5R + 5x, so x = R/3.
8. At what altitude above the Earth's surface will the gravitational force on a body be one-sixteenth of its value at the surface?
A.
R
B.
2R
C.
3R
D.
4R
Correct Answer:
3R
Explanation:
Setting F/F0 = 1/16 gives [R/(R+h)]² = 1/16. Thus R/(R+h) = 1/4, so R+h = 4R and h = 3R.
9. The Earth's radius is 6400 km and the surface value of gravitational acceleration is 9.8 m/s². What is the gravitational acceleration at an altitude of 1600 km?