请教两道高一物理题,谢谢!!!

2024-11-06 17:48:38
推荐回答(2个)
回答1:

1、万有引力提供向心力:
GMm/r^2=m4兀^2r/T^2 又有黄金代换GM=gR^2
r=(gR^2 T^2/4兀^2)^1/3
v=w(omiga)r=2兀/T*r可得
2、第二个物体开始下落时,第一个物体的竖直速度vy=gt=10m/s,以后,1物体和2物体以相同加速度运动,即1相对2无加速度,以10m/s匀速下落,而1、2的水平距离为100m,当1下落到地面时,时间T=(2h/g)^1/2=20s,此时1、2的竖直距离最大,为20*10=200m,而水平又距离100m,所以此时距离为(200^2+100^2)^1/2=100根号5 m

回答2:

1 根据向心力=万有引力 mMG/r^2 = 4pai^2 mr/T^2 得MG=4pai^2 r^3/T2
又知mMg/R^2 =mv^2/r 得MG=rv^2
所以合并两式得v=2pai r/T

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