Ta có :

\(\frac{P_h}{P_đ}=\frac{g_h}{g_đ}=\left(\frac{R}{R+h}\right)^2\)

\(\Rightarrow\frac{1}{9}=\left(\frac{R}{R+h}\right)^2\)

\(\Rightarrow\frac{1}{3}=\frac{R}{R+h}\)

\(\Rightarrow h=2R=6400\cdot2=12800\left(\text{km}\right)\)

Ta có :

\(S=v_0t+\frac{1}{2}at^2\)

\(\Rightarrow\text{ }2=0+\frac{1}{2}a\cdot2^2\)

\(\Rightarrow\text{ }a=1\text{ m/s}^2\)

\(m=\frac{F}{a}=\frac{2}{1}=2\text{ kg}\)

Ta có :

a) Trái Đất : \(\text{P}_{\text{TĐ}}=\text{mg}_{\text{TĐ}}=75\cdot9,8=735\text{N}\)

b) Mặt Trăng : \(\text{P}_{\text{MT}}=\text{mg}_{\text{MT}}=75\cdot1,7=127,5\text{N}\)

c) Sao Kim : \(\text{P}_{\text{SK} }=\text{mg}_{\text{SK}}=75\cdot8,7=652,5\text{N}\)

Ta có :

\(\frac{\text{P}_đ}{\text{P}_{\text{h}}}=\frac{\text{mg}_đ}{\text{mg}_{\text{h}}}=\frac{\text{g}_đ}{\text{g}_{\text{h}}}=\frac{\text{GM}\div\text{R}^2}{\text{GM}\div\left(\text{R}+\text{h}\right)^2}=\left(\frac{\text{R}+\text{h}}{\text{R}}\right)^2\)

\(\Rightarrow\frac{20}{10}=\left(\frac{6400+\text{h}}{6400}\right)^2\)

\(\Rightarrow\sqrt{2}=\frac{6400+\text{h}}{6400}\)

\(\Rightarrow\text{h}=2651\text{ km}\)

Giải :

\(\bigtriangleup\text{l}_1=\text{l}_1-\text{l}_0=30,5-20=0,5\left(\text{cm}\right)\)

\(\text{m}=\text{m}_1+\text{m}_2=20+80=100\left(\text{g}\right)=5\text{m}_1\)

\(\Rightarrow \bigtriangleup\text{l}_2=5\bigtriangleup\text{l}_1=5\cdot0,5=2,5\left(\text{cm}\right)\)

\(\Rightarrow\text{ l}_2=\text{l}_0+\bigtriangleup\text{l}_2=30+2,5=32,5\left(\text{cm}\right)\)

ko đăng đc ảnh

HT

Sjjbfbds

gggggggggggggggggggggggggggggg

rất nhỏ so với chiều dài quỹ đạo