Physics question...I am new here and would appreciate any help, thanks?

Question

A 2400 kg satellite is in a circular orbit around a planet (NOT earth). The satellite travels with a constant speed of 6.67 X 10^3 m/s. The radius of the circular orbit is 8.92 X 10^6 m.

a) What is the acceleration of the satellite?
b) What is the orbital period of the satellite?
c) What is the magnitude of the gravitational force exerted on the satellite by the planet?

Answer 1

for acceleration=mv62/r
for orbital period of statellite use v=2*pi*r/t u h pi value,r radiu,v velocity u have
magnitude u can use f=ma and f=gmm/r^2
g=6.7*10 raise to power -11
m1=2400
m2=6*10 raise to power 24
r=8.92 * ten raise to power 6m
hope this will help
regards
ghulam ali
pakistan

Answer 2

a). Acceleration is just centripetal acceleration

a = m v^2 / r

b). Find circumference of circle with radius 8.92 x 10^6 m, this will give you the distance the satellite travels in one orbit. Finding period is trivial from this point.

c). Newton's second law

F = m a

Answer 3

A eco-friendly automobile is following a yellow automobile down an constrained-get right of entry to highway at a relentless velocity of seventy 5.a million mi/hr. The separation distance between the two vehicles is 34.4 feet. The yellow automobile slams on the brakes and decelerates at a fee of -7.ninety m/s/s. After a reaction of time of 0.308 seconds, the eco-friendly automobile starts decelerating at a fee of -7.22 m/s/s. what's the suitable separation distance (in m) between the vehicles as quickly as stopped? If the trailing automobile (eco-friendly automobile) 'passes' the lead automobile (yellow automobile), then enter a adverse fee for the respond. (Given: a million m/s = 2.24 mi/hr; 3.28 feet = a million m) seventy 5.a million mi/hr ÷ 2.24 mi/hr / a million m/s = 33.fifty 3 m/s enable’s start up the clock whilst the yellow automobile slams on the brakes. The preliminary place is the region of the eco-friendly automobile. eco-friendly automobile’s distance whilst the yellow automobile slams on the brakes, the eco-friendly automobile continues to be going 33.fifty 3 m/s for 0.308 s = 10.33 m. next the eco-friendly automobile decelerates at -7.22 m/s^2 to discover the gap the eco-friendly automobile will holiday till it stops use the equation under. (very final velocity)^2 – (preliminary velocity)^2 = 2 * a * d 0 – (33.fifty 3)^2 = 2 * -7.22 * d d = [(33.fifty 3)^2] ÷ 14.40 4 d = seventy seven.86 m If we start up the clock on the time whilst the yellow automobile slams on brakes, the eco-friendly automobile could have traveled seventy seven.86 m + 10.33 m = 88.19 m eco-friendly automobile’s entire distance = 88.19 m The yellow automobile’s distance is the 34.4 feet it is at the back of the eco-friendly automobile plus the gap it travels because it decelerates at 7.ninety m/s^2 from 33.fifty 3 m/s to 0 m/s. (very final velocity)^2 – (preliminary velocity)^2 = 2 * a * d 0 – (33.fifty 3)^2 = 2 * -7.ninety * d d = [(33.fifty 3)^2] ÷ 15.80 d = seventy one.sixteen 34.4 feet ÷ 3.28 feet / a million m = 10.40 9 m entire distance = seventy one.sixteen + 10.40 9 = 80 one.sixty 5 m Separation distance = 80 one.sixty 5 m – 88.19 m = -6.fifty 4 m If, you choose me to respond to your questions interior the destiny, request that I be one in each of your contacts. Then once you submit a query, i will get carry of it via utilising digital mail. otherwise I purely seek randomly for a query to respond to. thank you for the prospect to be of help