1 Extra Newtonβs Law of Gravitation Practice Problems - Answers 1. Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitational attraction between them. = ! ! ! = 6.67 10 !!! ! ! 200 500 25 ! = . !2. Two spherical objects have masses of 1.5 x 10 5 kg and 8.5 x 10 2 kg. Their centers are separated by a distance of 2500 m. Find the gravitational attraction between them. = ! ! ! = 6.67 10 !!! ! ! 1.5 10 ! 8.5 10 ! 2500 ! = . !3. Two spherical objects have masses of 3.1 x 10 5 kg and 6.5 x 10 3 kg. The gravitational attraction between them is 65 N. How far apart are their centers? = ! ! ! , β= ! ! = 6.67 10 !!! ! ! 3.1 10 ! 6.5 10 ! 65 = . !4. Two spherical objects have equal masses and experience a gravitational force of 25 N towards one another. Their centers are 36 cm apart. Determine each of their masses. = ! ! ! , β, ! = ! = ; = ! ! NOTE: r is given in cm and MUST be converted to meters!!! = ! = = 0.36 25 6.67 10 !!! ! ! = . 5. A 1 kg object is located at a distance of 6.4 x10 6 m from the center of a larger object whose mass is 6.0 x 10 24 kg. a. What is the size of the force acting on the smaller object? = ! ! ! = 6.67 10 !!! ! ! 6.0 10 !" 1 6.4 10 ! ! = . b. What is the size of the force acting on the larger object? Newtonβs Third Law β the forces are equal so the answer is 9.8 N.
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Extra Practice - Newtons Law of Universal Gravitation Answers
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Extra Newtonβs Law of Gravitation Practice Problems - Answers
1. Two spherical objects have masses of 200 kg and 500 kg. Their centers are separated by a distance of 25 m. Find the gravitational attraction between them.
πΉ = πΊπ!π!
π!=
6.67 π₯ 10!!!ππ!
ππ! 200 ππ 500 ππ
25 π ! = π.π π ππ!π π΅
2. Two spherical objects have masses of 1.5 x 105 kg and 8.5 x 102 kg. Their centers are
separated by a distance of 2500 m. Find the gravitational attraction between them.
14. Two spherical objects have equal masses and experience a gravitational force of 25 N towards one another. Their centers are 36 cm apart. Determine each of their masses.
16. Compute g at a distance of 4.5 x 107m from the center of a spherical object whose mass is 3.0 x
1023 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 3.0 π₯ 10!" ππ
(4.5 π₯ 10! π )!= π.ππππ
πππ
17. Compute g for the surface of the moon. Its radius is 1.7 x106 m and its mass is 7.4 x 1022 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 7.4 π₯ 10!! ππ
(1.7 π₯ 10! π )!= π.π
πππ
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18. Compute g for the surface of a planet whose radius is twice that of the Earth and whose mass is the same as that of the Earth.
The radius is now 2RE (double the radius of the Earth). The radius has increased by a factor of two. If the radius increases by a factor of two then g decreases by a factor of 4 because g is inversely proportional the square of the radius.
g at the surface of the Earth is 9.8 m/s2
9.8 ππ !4
= π.ππ πππ
19. Compute g for the surface of the sun. Its radius is 7.0 x108 m and its mass is 2.0 x 1030 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 2.0 π₯ 10!" ππ
(7.0 π₯ 10! π )!= πππ
πππ
20. Compute g for the surface of Mars. Its radius is 3.4 x106 m and its mass is 6.4 x 1023 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 6.44 π₯ 10!" ππ
(3.4 π₯ 10! π )!= π.π
πππ
21. Compute g at a height of 6.4 x 106 m (RE) above the surface of Earth.
Need to first determine the r to use. It will be the radius of the Earth plus the height above the Earth. The height above the Earth is the same as the radius of the Earth. Thus, the distance separating the objects is 2RE. The radius has increased by a factor of two. If the radius increases by a factor of two then g decreases by a factor of 4 because g is inversely proportional the square of the radius.
g at the surface of the Earth is 9.8 m/s2
9.8 ππ !4
= π.ππ πππ
22. Compute g at a height of 2 RE above the surface of Earth.
The radius is now 2RE + RE which equals 3RE. The radius has increased by a factor of three. If the radius increases by a factor of three then g decreases by a factor of 9 because g is inversely proportional the square of the radius.
g at the surface of the Earth is 9.8 m/s2
9.8 ππ !9
= π.π πππ
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23. Compute g for the surface of a planet whose radius is half that of the Earth and whose mass is double that of the Earth.
The radius is now Δ RE. The radius has decreased by a factor of one-half. This increases g by a factor of 4 because g is inversely proportional to the square of the radius. The mass is now 2ME. The mass has increased by a factor of 2. This increases g by a factor of 2 because g is directly proportional to g. These two factors are multiplied together. Thus, g is increased by a factor of 8.
g at the surface of the Earth is 9.8 m/s2
9.8
ππ !β 8 = ππ.π
πππ
24. Compute g at a distance of 8.5 x 109m from the center of a spherical object whose mass is 5.0 x 1028 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 5.0 π₯ 10!" ππ
(8.5 π₯ 10! π )!= π.πππ
πππ
25. Compute g at a distance of 7.3 x 108 m from the center of a spherical object whose mass is 3.0 x
1027 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 3.0 π₯ 10!" ππ
(7.3 π₯ 10! π )!= π.ππ
πππ
26. Compute g for the surface of Mercury. Its radius is 2.4 x106 m and its mass is 3.3 x 1023 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 3.3 π₯ 10!" ππ
(2.4 π₯ 10! π )!= π.π
πππ
27. Compute g for the surface of Venus. Its radius is 6.0 x106 m and its mass is 4.9 x 1024 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 4.9 π₯ 10!" ππ
(6.0 π₯ 10! π )!= π.π
πππ
28. Compute g for the surface of Jupiter. Its radius of is 7.1 x107 m and its mass is 1.9 x 1027 kg.
π = πΊππ!
= 6.67 π₯ 10!!!ππ
!
ππ! 1.9 π₯ 10!" ππ
(7.1 π₯ 10! π )!= ππ
πππ
29. Compute g at a height of 4 RE above the surface of Earth.
The radius is now 4RE + RE which equals 5RE. The radius has increased by a factor of five. If the radius increases by a factor of five then g decreases by a factor of 25 because g is inversely proportional the square of the radius.
g at the surface of the Earth is 9.8 m/s2
9.8 ππ !25
= π.ππ πππ
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30. Compute g at a height of 5 RE above the surface of Earth. The radius is now 5RE + RE which equals 6RE. The radius has increased by a factor of six. If the radius increases by a factor of five then g decreases by a factor of 36 because g is inversely proportional the square of the radius.
g at the surface of the Earth is 9.8 m/s2
9.8 ππ !36
= π.ππ πππ
31. Compute g for the surface of a planet whose radius is double that of the Earth and whose mass is also double that of the Earth.
The radius is increased by a factor of 2. This decreases g by a factor of 4 because g is inversely proportional to the square of the radius. Double the mass then g increases by a factor of 2. These two factors are multiplied together. Thus, g is decreased by a factor of 2.