At a given point above Earthʹs surface, the acceleration due to gravity is equal to 7.8 m/s2. What is
the altitude of this point above Earthʹs surface? (G = 6.67 × 10-11 N · m2/kg2, Mearth = 5.97 × 1024
kg, Rearth = 6.38 × 106 m)
A) 1500 km
B) 2000 km
C) 2400 km
D) 970 km
E) 770 km
At a given point above Earthʹs surface, the acceleration due to gravity is equal to 7.8 m/s2......?
2016-06-10 3:43 am
回答 (2)
2016-06-17 8:26 am
✔ 最佳答案
g=MG/r^2 r=[(MG)/g]^1/2
h=r-R=[(MG)/g]^1/2-R
h=[(5.97x10^24*6.67x10^-11)/7.8]^1/2-6....
h=765 km ~ 770 km [E]
2016-06-10 4:05 am
G (Mearth m) / (Rearth + h) = m g
G Mearth / (Rearth + h) = g
(6.67 × 10⁻¹¹ N•m²/kg²) × (5.97 × 10²⁴ kg) / ([(6.38 × 10⁶ m) + h]² = 7.8 m/s²
[(6.38 × 10⁶ m) + h]² = (6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8 m²
(6.38 × 10⁶ m) + h = √[(6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8] m
h = √[(6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8] - (6.38 × 10⁶) m
h = 7.7 × 10⁵ m
h = 770 km
The answer is : E) 770 km
G Mearth / (Rearth + h) = g
(6.67 × 10⁻¹¹ N•m²/kg²) × (5.97 × 10²⁴ kg) / ([(6.38 × 10⁶ m) + h]² = 7.8 m/s²
[(6.38 × 10⁶ m) + h]² = (6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8 m²
(6.38 × 10⁶ m) + h = √[(6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8] m
h = √[(6.67 × 10⁻¹¹ ) × (5.97 × 10²⁴) / 7.8] - (6.38 × 10⁶) m
h = 7.7 × 10⁵ m
h = 770 km
The answer is : E) 770 km
收錄日期: 2021-04-18 14:57:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160609194347AAgAIc9