Условие задачи [ править ]
Определить работу (в джоулях), совершаемую при подъеме спутника с поверхности Земли на высоту
H
{\displaystyle H}
км. Масса спутника равна
m
{\displaystyle m}
т, радиус Земли
R
3
=
6380
{\displaystyle R_{3}=6380}
км. Ускорение свободного падения
g
{\displaystyle g}
у поверхности Земли положить равным 10 м/с2 .
m
=
7
,
0
{\displaystyle m=7{,}0}
т,
H
=
250
{\displaystyle H=250}
км.
По определению элементарная работа
d
A
=
F
(
x
)
d
x
{\displaystyle dA=F(x)dx}
, где
F
(
r
)
=
G
m
⋅
M
r
2
;
G
=
6
,
67
⋅
10
−
11
{\displaystyle F(r)=G{\frac {m\cdot M}{r^{2}}};G=6,67\cdot 10^{-11}}
Н*м*м / (кг*кг)
F
0
=
G
m
⋅
M
R
2
=
m
g
{\displaystyle F_{0}=G{\frac {m\cdot M}{R^{2}}}=mg}
F
x
=
G
m
⋅
M
(
R
+
x
)
2
{\displaystyle F_{x}=G{\frac {m\cdot M}{(R+x)^{2}}}}
-сила притяжения на высоте
x
{\displaystyle x}
F
x
F
0
=
G
m
⋅
M
R
2
G
m
⋅
M
(
R
+
x
)
2
=
R
2
(
R
+
x
)
2
⇒
F
x
=
F
0
R
2
(
R
+
x
)
2
=
m
g
R
2
(
R
+
x
)
2
=
m
g
(
1
+
x
R
)
2
{\displaystyle {\frac {F_{x}}{F_{0}}}={\frac {G{\frac {m\cdot M}{R^{2}}}}{G{\frac {m\cdot M}{(R+x)^{2}}}}}={\frac {R^{2}}{(R+x)^{2}}}\Rightarrow F_{x}={\frac {F_{0}R^{2}}{(R+x)^{2}}}={\frac {mgR^{2}}{(R+x)^{2}}}={\frac {mg}{\left(1+{\frac {x}{R}}\right)^{2}}}}
F
(
r
)
=
m
g
(
1
+
r
R
)
2
{\displaystyle F(r)={\frac {mg}{\left(1+{\frac {r}{R}}\right)^{2}}}}
A
=
∫
0
H
F
(
x
)
d
x
=
∫
0
H
m
g
(
1
+
x
R
)
2
d
x
=
7000
⋅
10
∫
0
250000
d
x
(
1
+
x
6380000
)
2
=
7000
⋅
10
⋅
6380000
∫
0
250000
d
(
x
6380000
+
1
)
(
x
6380000
+
1
)
2
=
{\displaystyle A=\int \limits _{0}^{H}F(x)dx=\int \limits _{0}^{H}{\frac {mg}{\left(1+{\frac {x}{R}}\right)^{2}}}dx=7000\cdot 10\int \limits _{0}^{250000}{\frac {dx}{\left(1+{\frac {x}{6380000}}\right)^{2}}}=7000\cdot 10\cdot 6380000\int \limits _{0}^{250000}{\frac {d\left({\frac {x}{6380000}}+1\right)}{\left({\frac {x}{6380000}}+1\right)^{2}}}=}
=
−
7000
⋅
10
⋅
6380000
⋅
1
x
6380000
+
1
|
0
250000
=
−
7000
⋅
10
⋅
6380000
(
1
250000
6380000
+
1
−
1
0
6380000
+
1
)
≈
1
,
68
⋅
10
10
{\displaystyle =-7000\cdot 10\cdot 6380000\cdot {\frac {1}{{\frac {x}{6380000}}+1}}|_{0}^{250000}=-7000\cdot 10\cdot 6380000\left({\frac {1}{{\frac {250000}{6380000}}+1}}-{\frac {1}{{\frac {0}{6380000}}+1}}\right)\approx 1,68\cdot 10^{10}}
(Дж)