题目内容
按万有引力定律,两质点间的吸引力F=k
,k为常数,m1,m2为两质点的质量,r为两点间距离,若两质点起始距离为a,质点m1沿直线移动至离m2的距离为b处(b>a),则克服吸引力所作之功为______.
| m1m2 |
| r2 |
克服吸引力所作之功为
W=
k
dr(-km1m2
)
=km1m2(
-
)
故答案为:km1m2(
-
)
W=
| ∫ | ba |
| m1m2 |
| r2 |
| 1 |
| r |
| | | ba |
| 1 |
| a |
| 1 |
| b |
故答案为:km1m2(
| 1 |
| a |
| 1 |
| b |
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