Question

The potential energy of a satellite, having mass
m and rotating at a height of 6.4 × 10⁶ m from the
earth surface, is
(a) – mgRₑ (b) – 0.67 mgRₑ
(c) – 0.5 mgRₑ (d) – 0.33 mgRₑ

Answer 1

Explanation:

just to compare the relation between g &G

Answer 2

Explanation:

We Should Know Some Trignometric Identities For Solving This Question.

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$\begin{gathered}1. \: \: \sin(2 \alpha ) = 2 \sin( \alpha ) \cos( \alpha ) \\\end{gathered} </p><p>1.sin(2α)=2sin(α)cos(α)$

$2. \: \: \sin(180 - \alpha ) = \sin( \alpha )2.sin(180−α)=sin(α)$

$3. \: \: \sin(90 - \alpha ) = \cos( \alpha )3.sin(90−α)=cos(α)$

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$L.H.S :- Sin10. Sin30. Sin50. Sin70$

$\begin{gathered}= \cos(90 - 10) \sin(30) \cos(90 - 40) \cos(90 - 70) \\ = \cos(80) \: \frac{1}{2} \: \cos(40) \: \cos(20) \\ = \frac{1}{4 \sin(20) } \cos(8 0 ) \cos(40) . \: 2 \sin(20) \cos(20) \\ = \frac{1}{8 \sin(20) } \cos(80) 2. \cos(40) \sin(40) \\ = \frac{1}{16 \sin(20) } 2 \cos(80) \sin(80 ) \\ = \frac{1}{16 \sin(20) } \sin(160) \\ = \frac{1}{16 \sin(20) } \sin(180 - 20) \\ = \frac{1}{16 \sin(20) } \sin(20) \\ = \frac{1}{16} \: \: \: \: \: \: \: \: .........R.H.S\end{gathered}$

$=cos(90−10)sin(30)cos(90−40)cos(90−70)$

$=cos(80) 2/1$

$cos(40)cos(20)= 4sin(20)1$

$cos(80)cos(40).2sin(20)cos(20)= 8sin(20)1$

$cos(80)2.cos(40)sin(40)= 16sin(20)1$

$2cos(80)sin(80)= 16sin(20)1$

$sin(160)= 16sin(20)1$

$sin(180−20)= 16sin(20)1$

$sin(20)= 16/1$

.........R.H.S,