Question

While visiting Planet Physics, you toss a rock straight up at 11 m/s and catch it 2.5 s later. While you visit the surface, your cruise ship orbits at an altitude equal to the planet’s radius every 230 min. What are the (a) mass and (b) radius of Planet Physics?

Short answer

a) Mass of the planet is 5.8 x 10 ²² kg

b) Radius of the planet is 1.36 x 10⁶ m

Step-by-step solution

Part(a) Step1: Given information

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

Part(a)Step2: Explanation

From Keplar's law we get period as
$T^2=\frac{4{\pi }^2}{GM}{r}^3$
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

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

From keeper's law

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

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

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

$R=\frac{g{T}^2}{32{\pi }^2}$

Substitute the values

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

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

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

Mass of planet is 5.8 x 10²² kg

Part(b)Step2:Given information

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

Part(b)Step2:Explanation

From Keplar's law we get period as

$T^2=\frac{4{\pi }^2}{GM}{r}^3$

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

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

Substitute the values

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

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