91影视

Speed of the rock v =11 m/s
Time to return, t=2.5 s
Time period of the orbit, T=230 min

From Keplar's law we get period as
$T^2=\frac{4{蟺}^2}{GM}{r}^3$
Where,
T= time period
G = universal gravitational constant
M = mass of planet
r = radius

Also from simple kinematics we know that the acceleration of the planet can be calculated as
v = u - g t
where t is time which is half the time to return.

Substitute the value and find the value of g, we get

$$\begin{gathered} 0=11m/s-g脳\frac{2.5s}{2} \\ g=8.8\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From keeper's law

$T^2=\frac{4{蟺}^2}{GM}{r}^3$
where r= 2R, so we get

$$\begin{gathered} T^2=\frac{32{蟺}^2{R}^3}{GM} \\ M=\frac{32{蟺}^2{R}^3}{G{T}^2}.......................................\left(1\right) \end{gathered}$$

We can also calculate M from gravitational acceleration formula as

localid="1649664539530" $M=\frac{g{R}^2}{G}............................................\left(2\right)$

equate (1) and (2) and solve for R, we get

$R=\frac{g{T}^2}{32{蟺}^2}$

Substitute the values

$$\begin{gathered} R=\frac{(8.8m/s^2)脳(230脳60s{)}^2}{32{蟺}^2} \\ =685213.1\mathrm{m} \end{gathered}$$

Now substitute the value of R in equation (2) to get value of mass

$$\begin{gathered} M=\frac{g{R}^2}{G} \\ =\frac{(8.8m/s^2)脳(685213.1m{)}^2}{(6.67脳{10}^{-11}N{m}^2/k{g}^2)} \\ =5.8脳{10}^{22}\mathrm{kg} \end{gathered}$$

Mass of planet is 5.8 x 10²² kg

Speed of the rock v =11 m/s
Time to return, t=2.5 s
Time period of the orbit, T=230 min

From Keplar's law we get period as

$T^2=\frac{4{蟺}^2}{GM}{r}^3$

Where,
T= time period
G = universal gravitational constant
M = mass of planet
r = radius

Radius of planet can be calculated as
$R=\frac{g{T}^2}{32{蟺}^2}$

Substitute the values

$$\begin{gathered} R=\frac{(8.8m/s^2)脳(230脳60s{)}^2}{32{蟺}^2} \\ =685213.1\mathrm{m} \end{gathered}$$

Radius of planet physics is r =2 R
r = 2 x 685213.1 m
r = 1.36 x 10⁶ m