我有一題關於自由落體的問題
2010-08-12 3:12 am
在一個半徑為r米的星球上(沒有空氣)(完美的球體)，在地平面上m米的一個非常小的物體由開始下跌至掉在地面上所用的時間是多少？(設在地面上的重力加速度為g m/s^2)(答案應該不是(2m/g)^0.5，因為在不同地方的重力加速度是不同的。距離這個星球的中心越遠，重力加速度越小。)
回答 (3)
2010-08-13 1:56 am
✔ 最佳答案
對不起!缺乏足夠資料而無法計算。
如果運用「newton's law of gravitation」，
F = G [(mass of 星球 * mass of 非常小的物體) / r^2]
F is gravitational force
G is universal gravitational constant
r is the distance between centres of both object
therefore ，地面上的重力加速(g) = G [mass of 星球 / (r+m)^2] ，
再用番 s = 0.5 (g)( t^2) ，
t ={ [2m(r+m)^2] / G(mass of 星球) }^0.5
參考: new way physics for advanced level -mechanics Mok Tim-ming , Wong Chi-san , Poh Liong-yong
2010-08-13 11:09 pm
2010-08-13 10:33 pm
First find the velocity v of the falling object at distance h above the planet surface.
From conservation of mechanical energy,
(1/2)(mo)v^ = GM(mo).[1/(r+h) - 1/(r+m)]
where (mo) is the mass of the object, M is the mass of the planet and G is the Universal Gravitational Constant.
we then get, v = square-root[2GM.(1/(r+h) - 1/(r+m))]
since v = dh/dt
hence, dh/dt = square-root[2GM.(1/(r+h) - 1/(r+m))]
i.e. dh/square-root[2GM.(1/(r+h) - 1/(r+m))] = dt
You need to integrate the two sides, the integration limit of the left hand side is from h = m to h = 0, whereas the integration limit for the right hand side is from t = 0 to t = T, say (where T is the time that the object reach the surface).
Therefore T = intergral{ dh/square-root[2GM.(1/(r+h) - 1/(r+m))] }, with limt from h = m to h = 0
The integration is not easy to perform. Perhaps you need to evaluate it by numerical integration.
From conservation of mechanical energy,
(1/2)(mo)v^ = GM(mo).[1/(r+h) - 1/(r+m)]
where (mo) is the mass of the object, M is the mass of the planet and G is the Universal Gravitational Constant.
we then get, v = square-root[2GM.(1/(r+h) - 1/(r+m))]
since v = dh/dt
hence, dh/dt = square-root[2GM.(1/(r+h) - 1/(r+m))]
i.e. dh/square-root[2GM.(1/(r+h) - 1/(r+m))] = dt
You need to integrate the two sides, the integration limit of the left hand side is from h = m to h = 0, whereas the integration limit for the right hand side is from t = 0 to t = T, say (where T is the time that the object reach the surface).
Therefore T = intergral{ dh/square-root[2GM.(1/(r+h) - 1/(r+m))] }, with limt from h = m to h = 0
The integration is not easy to perform. Perhaps you need to evaluate it by numerical integration.
收錄日期: 2021-04-23 21:52:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100811000051KK01514