A2 Centripetal Force/Acceleration question

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jf1994
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'The earth moves round the Sun on a circular orbit of radius 1.5x10^11 m, taking 365.25 days for each orbit. Calculate:
a) the speed
b) the centripetal acceleration of the earth on its orbit around the sun.

Part B

A satellite is in orbit just above the surface of a spherical planet which has the same radius as the earth and the same acceleration of freefall at its surface. Calculate:
a) the speed
b) the time taken for 1 complete orbit of this satellite.

Radius of the earth = 6400km, Acceleration of free fall = 9.8 ms^-2'

I've done part A and got v = 3.0x10^4 ms^-1 and a = 5.9x10^-3 ms^-2 but I can't figure out which equation to use to do the next part with the information given. Can anyone point me in the right direction? Thanks
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Stonebridge
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(Original post by jf1994)
'The earth moves round the Sun on a circular orbit of radius 1.5x10^11 m, taking 365.25 days for each orbit. Calculate:
a) the speed
b) the centripetal acceleration of the earth on its orbit around the sun.

Part B

A satellite is in orbit just above the surface of a spherical planet which has the same radius as the earth and the same acceleration of freefall at its surface. Calculate:
a) the speed
b) the time taken for 1 complete orbit of this satellite.

Radius of the earth = 6400km, Acceleration of free fall = 9.8 ms^-2'

I've done part A and got v = 3.0x10^4 ms^-1 and a = 5.9x10^-3 ms^-2 but I can't figure out which equation to use to do the next part with the information given. Can anyone point me in the right direction? Thanks
As it orbits just above the surface the distance it travels in one orbit is just the circumference of the planet. You are told its radius is the same as Earth. Then it's just the standard simple relationship between time (for one orbit), distance (circumference) and speed (orbital speed as calculated in part a)
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Master219
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(Original post by jf1994)
'The earth moves round the Sun on a circular orbit of radius 1.5x10^11 m, taking 365.25 days for each orbit. Calculate:
a) the speed
b) the centripetal acceleration of the earth on its orbit around the sun.

Part B

A satellite is in orbit just above the surface of a spherical planet which has the same radius as the earth and the same acceleration of freefall at its surface. Calculate:
a) the speed
b) the time taken for 1 complete orbit of this satellite.

Radius of the earth = 6400km, Acceleration of free fall = 9.8 ms^-2'

I've done part A and got v = 3.0x10^4 ms^-1 and a = 5.9x10^-3 ms^-2 but I can't figure out which equation to use to do the next part with the information given. Can anyone point me in the right direction? Thanks
Hmm, i thought it would be:

Part a) the centripetal force acting on the satellite is gravity. hence the equation F=mv^2/r and F=GMm/r^2 can be equated. This leads to the m's being cancelled (mass of satellite, hence y ur not given it). M is the mass of the earth (if u do AQA then thats in the formula booklet). both r's are radius of orbit. v is the speed they r asking for. And F in the first equation is centripetal force and F in the second equation is force between the two masses. U then rearrange for v=... and plug in numbers.

Part b) well u have now calculated v. and u may be aware that w (angular speed, the wierd w)=v/r where r is the radius of the orbit and v is the speed we just calculated. and as wierd angular speed w = 2pi/T where T= time period therefore, 2pi/T=v/r u just rearrange for T and u can get the time for one cycle.

Let me know if u want me to clarify anything
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james153
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I believe you're actually looking for a combination of the two approaches above.

a) You're given acceleration of free fall and I suggest you use it.
Hint: use the relation F=mg where g is the acceleration of free fall
and the relation: F= \frac{mv^2}{r} to find the velocity in terms of the 2 constants you're given in the question.

b) Follow the approach as outlined above by Stonebridge, using the relationship between speed, distance and time.
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KeithHayward
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(Original post by jf1994)
'The earth moves round the Sun on a circular orbit of radius 1.5x10^11 m, taking 365.25 days for each orbit. Calculate:
a) the speed
b) the centripetal acceleration of the earth on its orbit around the sun.

Part B

A satellite is in orbit just above the surface of a spherical planet which has the same radius as the earth and the same acceleration of freefall at its surface. Calculate:
a) the speed
b) the time taken for 1 complete orbit of this satellite.

Radius of the earth = 6400km, Acceleration of free fall = 9.8 ms^-2'

I've done part A and got v = 3.0x10^4 ms^-1 and a = 5.9x10^-3 ms^-2 but I can't figure out which equation to use to do the next part with the information given. Can anyone point me in the right direction? Thanks
Have u got the answers for part a?
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Master219
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(Original post by james153)
I believe you're actually looking for a combination of the two approaches above.

a) You're given acceleration of free fall and I suggest you use it.
Hint: use the relation F=mg where g is the acceleration of free fall
and the relation F=mv2/r to find the velocity in terms of the 2 constants you're given in the question.

b) Follow the approach as outlined above by Stonebridge, using the relationship between speed, distance and time.
a) but the centripetal force, the force thats keeping the satellite in place is gravity, i.e. the force of attraction between the satellite and the planet, newtons law of gravitation. At first i thought mg cuz they giv us g but wt makes more sense, and the way i always have done these satellite qs from past papers (and checked mark schemes), is the way i quoted above...

(Original post by Stonebridge)
As it orbits just above the surface the distance it travels in one orbit is just the circumference of the planet. You are told its radius is the same as Earth. Then it's just the standard simple relationship between time (for one orbit), distance (circumference) and speed (orbital speed as calculated in part a)

The speed in part a is for earth around the sun.... but this Part B) a) is for sat around earth. So u hav to find the speed for the satellite around the earth first and then use that in the 2pi/T = v/r (as i mentioned above)
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james153
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(Original post by Master219)
a) but the centripetal force, the force thats keeping the satellite in place is gravity, i.e. the force of attraction between the satellite and the planet, newtons law of gravitation. At first i thought mg cuz they giv us g but wt makes more sense, and the way i always have done these satellite qs from past papers (and checked mark schemes), is the way i quoted above...
F=mg is another way of expressing Newton's law of gravitation.
Newton's law of gravitation is normally expressed F= \frac{GMm}{r^2}
however, g= \frac{GM}{r^2}.

Therefore although the method you propose is not incorrect, it involves using the mass of the earth instead of the acceleration due to gravity, and considering you're given acceleration it makes more sense to use that approach, and it's also a bit tidier.
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Stonebridge
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(Original post by Master219)


The speed in part a is for earth around the sun.... but this Part B) a) is for sat around earth. So u hav to find the speed for the satellite around the earth first and then use that in the 2pi/T = v/r (as i mentioned above)
Thanks but I was discussing part B b) in which the answer to part B a) was needed for part B b)
I didn't comment on part B a) as I had misread the original post, thinking the student had already done this part. When I discovered this, it was unnecessary to do any more as another poster had already answered that point.
Hope that's cleared up the misunderstanding.
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Master219
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(Original post by james153)
F=mg is another way of expressing Newton's law of gravitation.
Newton's law of gravitation is normally expressed F=GMm/r^2
however, g is equal to GM/r^2.

Therefore although the method you propose is not incorrect, it involves using the mass of the earth instead of the acceleration due to gravity, and considering you're given acceleration it makes more sense to use that approach, and it's also a bit tidier.
Oh yh lol forgot about that g= thing haha
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Master219
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(Original post by Stonebridge)
Thanks but I was discussing part B b) in which the answer to part B a) was needed for part B b)
I didn't comment on part B a) as I had misread the original post, thinking the student had already done this part. When I discovered this, it was unnecessary to do any more as another poster had already answered that point.
Hope that's cleared up the misunderstanding.
yh got it thanks

(Original post by james153)
F=mg is another way of expressing Newton's law of gravitation.
Newton's law of gravitation is normally expressed F=GMm/r^2
however, g is equal to GM/r^2.

Therefore although the method you propose is not incorrect, it involves using the mass of the earth instead of the acceleration due to gravity, and considering you're given acceleration it makes more sense to use that approach, and it's also a bit tidier.
but pretty much always when these questions come up they don't tell u g. Normally, it's a satellite around the earth and as the earth has a radial gravitational field the g where the satellite would b is different to the surface. But for this q u can just use the g they giv u, as u and the other poster mentioned. So OP my method is the general method that can be used on any similar q regardless of whether they giv u g or not, but the F=mg method (which is derived from the F=GMm etc formula) is used when they giv u g (although if u use my method ull get the same answer but as James correctly mentioned, it may get a bit untidy, especially if ur maths skills rnt the best)
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