At a certain time a star of mass 7 times 10^30 kg is located at m. A planet of m

Question

At a certain time a star of mass 7 times 10^30 kg is located at m. A planet of mass 3 times 10^24 kg is located at m and is moving with a velocity of m/s. What force is acting on the planet? At a time 1 times 10^6 seconds later, what is the new velocity of the planet? Show your work clearly. Where is the planet's positon at this later time? Show your work clearly. Explain briefly why the procedures you followed in parts (a) and (b) were able to produce usable results but wouldn't work if the later time had been 1 times 10^9 seconds instead of 1 times 10^6 seconds after the initial time.

Explanation / Answer

According to the universal gravitational law

F= Gm1m2/r^2

where m1=mass of star=7*10^30 kg

m2= mass of planet=3*10^24 kg

r= distance between the planet and the star=5*10^12 to 2*10^12m

Taking average of the distance=3.5*10^12m

G= gravitational constant=6.67*10^-11m3kg-1s-2

Force acting on the planet= F=Gm1m2/r^2

F=6.67*10-11* 7*10^30* 3*10^24/(3.5*10^12)^2

=140.07*10^30*10^24/12.25*10^24

=11.43*10^30N

Velocity= Distance travelled/ time taken

Initially velocity of planet=0.3 *10^4m/s

Initial distance between star and planet= 5*10^12m

Initial time taken=Initial distance/ initial speed

=5*10^12/0.3*10^4

=16.6*10^8s

Now after the 1*10 sec later

16.6*10^8s+1*10^6 s

=10^6(16.6*10^2+1)

=10^6(16.6*10^2+1*10^2) ( since 1*10^2= 1 only no change in value)

=10^6*17.6*10^2=17.6*10^8s

New velocity= New distance/time taken

=2*10^12/17.6*10^8

=0.113*10^4m/s

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