hard question... (1 Viewer)

[supsations]

this is a question that is hard
can u help me do it???
THANKS!

A satellite of mass 150 kg orbits Earth at a distance of 7.8x10^6 m (from centre of earth)

1) calculate the magnitude of gravitationsal force between EArth and the satellite

2) the magnitude of the centripetal acceleration of the satellite

3) the speed of the satellite in its orbit

4) GPE in relation to the Earth

5) velocity at which the satellite would crash into the earth.

 

[KFunk]

this is a question that is hard
can u help me do it???
THANKS!

A satellite of mass 150 kg orbits Earth at a distance of 7.8x10^6 m (from centre of earth)

1) calculate the magnitude of gravitationsal force between EArth and the satellite

2) the magnitude of the centripetal acceleration of the satellite

3) the speed of the satellite in its orbit

4) GPE in relation to the Earth

5) velocity at which the satellite would crash into the earth.

1) F_G = Gm_em_s/r², I'll assume that G = 6.67 x 10^-11 and that m_e = 5.98 x 10²⁴ kg
---] F_G = 983.4 N

2) F_C = F_G
a_c = a_g = 983.4/150 = 6.556 ms^-2

3) a_c = v²/r = 6.556 ms^-2
----] v = √(6.556 ms^-2x7.8x10⁶ m) = 7,151 ms^-1

4) E_p = -Gm_em_s/r = -7.6705 x 10<sup>8 J = -7670.5 MJ

5) I'm going to think of how to do this one. I expect you would need to use integration to do it properly since you're not dealing with uniformly accelerated motion.

(Edit: Fixed up the Superscript and subscript codes for you. - Xayma)</sup>

 

[supsations]

hey...

u seem to have html tags in your answer...
wats </sub> anyawy?

 

[KFunk]

Bah, it's not enabled in the physics forum yet. Sub is for subtext. I'll quickly neaten up but I gotta go have dinner.

 

[supsations]

thanks anyways

nah its ok

i just compiled it using my worthy notepad

 

[KFunk]

1) F_G = Gmm/r^2, I'll assume that G = 6.67 x 10^-11and that m_e = 5.98 x 10^24 kg
---> F_G = 983.4 N

2) F_C= F_G
a_c = a_g = 983.4/150 = 6.556 ms^-2

3) a_c = v^2r = 6.556 ms^-2
----> v = sqrt(6.556 ms^-2 x7.8x10^6 m) = 7,151 ms^-1

4) E_p = -Gmmr = -7.6705 x 10^8 J = -7670.5 MJ

5) I'm going to think of how to do this one. I expect you would need to use integration to do it properly since you're not dealing with uniformly accelerated motion.

 

[supsations]

worked it ou

i think i worked it out

last question) GPE at satelllite= -7.7 x 10^31
GPE on Earth: -9.4 x 10^34

negate the 2 : 9.37 x 10^34


KE=GPE
1/2 m v^2 = 9.37 x 10^34

v = 3.5 x 10^16

 

[KFunk]

I have a sneaking suspicion that the satellite wont reach the earth's surface at a speed of 10^8 times c . A speed close to that of light would be unlikely in itself, but what you've got is something else altogether .

 

[KFunk]

i think i worked it out

last question) GPE at satelllite= -7.7 x 10^31
GPE on Earth: -9.4 x 10^34

negate the 2 : 9.37 x 10^34


KE=GPE
1/2 m v^2 = 9.37 x 10^34

v = 3.5 x 10^16

I'm not sure how you got those values but remember that:
Gravitational potential energy = -GMm/r
where G = 6.67 x 10^-11, M= 5.98 x 10^24, m= mass of the object and r= distance from the earth.

GPE at distance of 7.8x10^6 m = -7.6705 x 10^9 J (my bad, I think I said 10^8 above)
GPE at 6.38 x 10^6 m (radius) = -9.3777 x 10^9 J

The difference = 1.7 x 10^9 J
Using your logic 1/2.mv^2 = 1.7 x 10^9 J ---> v = sqrt[(2x1.7 x 10^9 J)/150] = 4,771 ms^-1 which is much more plausible. Mind you, a lot of the energy would be converted into heat/sound etc during entry into the atomsphere.

 

[supsations]

thanks

thanks dancing bear person
mind u ... that was in an exam ... and it was the 2nd question

 

[KFunk]

Just because it's question two doesn't mean it should be easy . Don't stress about it. The question I disagreed with most on my chemistry exam was question 1... and it was multiple choice!