Centripetal force exerted on Earth

[electronicxco]

Here is the problem and the choices:
Calculate the centripetal force exerted on
the Earth by the Sun. Assume that the period
of revolution for the Earth is 365.25 days,
the average distance is 1.5 × 10^8 km and the
Earth’s mass is 6 × 10^24 kg.
The choices are:
1. 3.56775 × 10^22 N
2. 2.66331 × 10^32 N
3. 7.24562 × 10^22 N
4. 1.62932 × 10^21 N
5. None of these
6. 3.56775 × 10^19 N
7. 4.6238 × 10^29 N
8. 1.28439 × 10^26 N
9. 7.24562 × 10^20 N
__________________________________________________
Here is the attempted solution ( got choice # 6 but it is not correct)

2πr=D
(2)(π)(1.5 * 10^8) = 942477796.1 m=D

D = VT
V = D/T
V = (942477796.1) / ( 365.25 * 24 * 60 * 60)
V = 29.86 m/s


F = MV^2 / R
F = (6 * 10^24)(29.86)^2 / (1.5 * 10^8)
F = 3.56775 * 10^19

 

[PhanthomJay]

Here is the problem and the choices:
Calculate the centripetal force exerted on
the Earth by the Sun. Assume that the period
of revolution for the Earth is 365.25 days,
the average distance is 1.5 × 10^8 km and the
Earth’s mass is 6 × 10^24 kg.
The choices are:
1. 3.56775 × 10^22 N
2. 2.66331 × 10^32 N
3. 7.24562 × 10^22 N
4. 1.62932 × 10^21 N
5. None of these
6. 3.56775 × 10^19 N
7. 4.6238 × 10^29 N
8. 1.28439 × 10^26 N
9. 7.24562 × 10^20 N
__________________________________________________
Here is the attempted solution ( got choice # 6 but it is not correct)

2πr=D
(2)(π)(1.5 * 10^8) = 942477796.1 m=D

that ditance is in km, not m

D = VT
V = D/T
V = (942477796.1) / ( 365.25 * 24 * 60 * 60)
V = 29.86 m/s


F = MV^2 / R
F = (6 * 10^24)(29.86)^2 / (1.5 * 10^8)
F = 3.56775 * 10^19

looks like your off by a factor of 1000. Simple units error.

 

[electronicxco]

wow..thanks for catching my mistake!