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

Expert verified

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

01

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

02

Part(a)Step2: Explanation

From Keplar's law we get period as
T2=4π2GMr3
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

0=11m/s-g×2.5s2g=8.8m/s2

From keeper's law

T2=4π2GMr3
where r= 2R, so we get

T2=32π2R3GMM=32π2R3GT2.......................................(1)

We can also calculate M from gravitational acceleration formula as

localid="1649664539530" M=gR2G............................................(2)

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

R=gT232π2

Substitute the values

R=(8.8m/s2)×(230×60s)232π2=685213.1m

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

M=gR2G=(8.8m/s2)×(685213.1m)2(6.67×10-11Nm2/kg2)=5.8×1022kg

Mass of planet is 5.8 x 1022 kg

03

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

04

Part(b)Step2:Explanation

From Keplar's law we get period as

T2=4π2GMr3

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

Radius of planet can be calculated as
R=gT232π2

Substitute the values

R=(8.8m/s2)×(230×60s)232π2=685213.1m

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

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