这道关于外有引力定律物理题怎么做
答案是AB。
mv²/(R+h)=mGM/(R+h)²,
v²=GM/(R+h)=R²GM/[R²(R+h)]=R²(GM/R²)/(R+h)=R²g/(R+h)
∴速率v=R[g/(R+h)]^(½)
周期T=2π(R+h)/v=2π(R+h)/{R[g/(R+h)]^(½)}=2π[(R+h)/R][(R+r)/g]^(½)
加速度a=F/m=[mGM/(R+h)²]/m=GM/(R+h)²=GMR²/[R²(R+h)&sup...全部
答案是AB。
mv²/(R+h)=mGM/(R+h)²,
v²=GM/(R+h)=R²GM/[R²(R+h)]=R²(GM/R²)/(R+h)=R²g/(R+h)
∴速率v=R[g/(R+h)]^(½)
周期T=2π(R+h)/v=2π(R+h)/{R[g/(R+h)]^(½)}=2π[(R+h)/R][(R+r)/g]^(½)
加速度a=F/m=[mGM/(R+h)²]/m=GM/(R+h)²=GMR²/[R²(R+h)²]
=(GM/R²)[R²/(R+h)²]=g[R²/(R+h)²]
重力势能Ep这样计算：设想飞船从离地球距离为r处移动到无穷远，这要克服引力做负功，势能增加，
Ep(无穷远)-Ep(r)=-∫(mGM/r²)dr，
=-mGM/r(无穷远)-(-mGM/r)=mGM/r，
其中，无穷远处Ep(无穷远)=-mGM/r(无穷远）=0
∴Ep(r)=-mGM/r
r=R+h，Ep(r)=-mGM/(R+h)=-mGMR²/[R²(R+h)]
=-m(GM/R²)R²/(R+h)=-mgR²/(R+h)。
。收起
