Gravitation

Initial energy ($E_i$): The satellite is at rest on Earth’s surface ($r = R$), so

$$E_i \;=\; U_i \;=\; -\,\frac{G\,M\,m}{R}.$$

Final energy ($E_f$): The satellite is in a circular orbit at radius $r = 3R$ (Earth’s radius $R$ plus altitude $2R$). For a circular orbit,

$$E_f \;=\; -\,\frac{G\,M\,m}{2\,(3R)} \;=\; -\,\frac{G\,M\,m}{6\,R}.$$

Minimum energy required = $\Delta E = E_f - E_i$:

$$\Delta E \;=\; -\,\frac{G\,M\,m}{6\,R} \;-\; \Bigl(-\,\frac{G\,M\,m}{R}\Bigr) \;=\; \frac{G\,M\,m}{R} \Bigl(1 \;-\;\tfrac{1}{6}\Bigr) \;=\; \frac{5\,G\,M\,m}{6\,R}.$$

$\text{Required energy\hspace{0.25em}\hspace{0.05em}}=\frac{5GMm}{6R}\mathrm{.}$

Escape velocity v = √(2GM/R).

Earth: v = √(2GMₑ/Rₑ)

Planet: M' = 9Mₑ, R' = 16Rₑ.

v' = √(2G(9Mₑ)/(16Rₑ)) = √(9/16) * √(2GMₑ/Rₑ) = (3/4)v

• Escape velocity v = √(2GM/R).
• Earth: v = √(2GMₑ/Rₑ). Planet: M' = 9Mₑ, R' = 16Rₑ. v' = √(2G(9Mₑ)/(16Rₑ)) = √(9/16) * √(2GMₑ/Rₑ) = (3/4)v.

(NEW NCERT 11^th Page No. 135, 136)

Assuming the Sun to be a solid sphere, \(I =(2 / 5) m R ^2\) Using conservation of angular momentum, \(I^{\prime} \omega\) ' \(=I \omega\)

\(\begin{array}{l}\Rightarrow(2 / 5) m (2 R)^2 \times\left(2 \pi / T ^{\prime}\right)=(2 / 5) mR ^2 \times(2 \pi / T ) \\ \Rightarrow T ^{\prime}=4 T=4 \times 27= 1 0 8 \text { days }\end{array}\)

mass of planet $\frac{M}{10}$ (M = mass of earth)

Radius of planet $=\frac{R}{2}$ (R = radius of earth)

Acceleration due to gravity on planet is

$g=\frac{GM/10}{{(r/2)}^2}=\frac{2}{5}\frac{GM}{R^2}=0.4\times g=0.4\times 9.8$

$=3.92 m/s^2$