91Ó°ÊÓ

r= radius of the orbit =25 x 10³m
T = time period = 11 days =(11)x(24)x(60)x(60) s =950400 s

Velocity can be calculated by using the formula

$v=\frac{2×\mathrm{π}×r}{T}..............................\left(1\right)$

Substituting all the values in eqation(1), we get

$$\begin{gathered} v=\frac{2×3.14×25×{10}^3}{950400} \\ v=0.16{\mathrm{ms}}^{-1} \end{gathered}$$

Velocity is 0.16 m/s

r= radius of the orbit =25 x 10³m
T = time period = 11 days =(11)x(24)x(60)x(60) s =950400 s

For the satellite revolution in stable condition,

The centripetal force of satellite = Gravitational force between satellite and the comet

$\frac{m{v}^2}{r}=\frac{GMm}{r^2}...................................\left(1\right)$
Where

M= mass of the comet
m= mass of the satellite
r= radius of the orbit
G= universal gravitational constant.

Simplify equation(1)

From equation (2) and (3) find the value of M

$$\begin{gathered} ⇒M=\frac{v^{2}r}{G}................................\left(2\right) \\ And, \\ v=\frac{2×\mathrm{π}×r}{T}.............................\left(3\right) \end{gathered}$$

Now substitute values and calculate

$$\begin{gathered} M=\frac{2^2×(3.14{)}^2×{\left(25×{10}^{3}m\right)}^3}{(950400s{)}^2×\left(6.647×{10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2}\right)} \\ M=1.02×{10}^{13}\mathrm{kg} \end{gathered}$$

Mass is 1.02 x 10¹³ kg

R = radius of the comet = 1.8 x 10³ m
r= radius of the orbit = 25 x 10³m
t= time taken =7hr = (7 x (60)(60) s =25200 s
vi= initial speed of the satellite =70 cm/s =0.7 m/s.

From the basic kinematics we know

$v_l=v_i+\frac{GM}{(r+R{)}^2}t.........................\left(5\right)$

Initial velocity =vi
Final velocity =vl
Acceleration a=$\frac{GM}{(r+R{)}^2}$
Time taken= t
Substituting values in v=u+a t, we get

$$\begin{gathered} ⇒v_l=(0.7m/s)+\left(\frac{(6.647×{10}^{-11}{m}^{3}k{g}^{-1}{s}^{-2})×(1.02×{10}^{13}kg)×\left(25200s\right)}{{\left(25×{10}^{3}m+1.8×{10}^{3}m\right)}^2}\right) \\ ⇒v_l=0.84\mathrm{m}{\mathrm{s}}^{-1} \end{gathered}$$

Landing velocity is 0.84m/s