Drive the escape velocity. Obtain an expression for escape velocity of a body from the surface of earth
Answers
Answer:
The formula for escape velocity comprises of a constant, G, which we refer to as the universal gravitational constant. The value of it is = 6.673 × 10-11 N . m2 / kg2. The unit for escape velocity is meters per second (m/s).
answer : 441/113
given, tan\theta=\frac{1}{\sqrt{7}}=\frac{p}{b}tanθ=
7
1
=
b
p
so, p = 1 and b = √7
from Pythagoras theorem,
h² = p² + b² => h = √(1² + √7²) = 2√2
hence, cos\theta=\frac{b}{h}=\frac{\sqrt{7}}{2\sqrt{2}}cosθ=
h
b
=
2
2
7
so, sec\theta=\frac{2\sqrt{2}}{\sqrt{7}}secθ=
7
2
2
sin\theta=\frac{p}{h}=\frac{1}{2\sqrt{2}}sinθ=
h
p
=
2
2
1
so, cosec\theta=\frac{2\sqrt{2}}{1}cosecθ=
1
2
2
we have to find \frac{cosec^2\theta-sin^2\theta}{cos^2\theta+sec^2\theta}
cos
2
θ+sec
2
θ
cosec
2
θ−sin
2
θ
= \frac{(2\sqrt{2})^2-\left(\frac{1}{2\sqrt{2}}\right)^2}{\left(\frac{\sqrt{7}}{2\sqrt{2}}\right)^2+\left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2}
(
2
2
7
)
2
+(
7
2
2
)
2
(2
2
)
2
−(
2
2
1
)
2
= \frac{8-\frac{1}{8}}{\frac{7}{8}+\frac{8}{7}}
8
7
+
7
8
8−
8
1
= \frac{64-1}{8\frac{(49+64)}{56}}
8
56
(49+64)
64−1
= \frac{63\times7}{113}
113
63×7
= \frac{441}{113}
113
441