- #1

Matthew 289

- 2

- 0

## Homework Statement

Distance Earth-Sun at perihelion = 1.471×10

^{8}km

Distance Earth-Sun at aphelion = 1.521×10

^{8}km

Sun mass = 2.0×10

^{30}kg

Earth mass = 5.972×10

^{24}kg

G = 6.67×10

^{-11}m

^{3}/kg⋅s

What is the change in Newton of the attraction force between the Sun and the Earth from the perihelion to the aphelion ?

## Homework Equations

F= G × M

_{sun}×M

_{Earth}/distance

^{2}

## The Attempt at a Solution

F

_{perihelion}= (6.67×10

^{-11}m

^{3}/kg⋅s) × (2.0×10

^{30}kg)×(5.972×10

^{24}kg)/(1.471×10

^{8}km) = 3.68×10

^{28}N

F

_{aphelion}= (6.67×10

^{-11}m

^{3}/kg⋅s) × (2.0×10

^{30}kg)×(5.972×10

^{24}kg)/(1.521×10

^{8}km) = 3.44×10

^{28}N

ΔF= |F

_{perihelion}| - |F

_{aphelion}| = 3.68×10

^{28}N - 3.44×10

^{28}N = 2.4×10

^{27}N

**My problem:**

The result does not correspond to that given by the textbook (2.37×10

^{21}N).

I cannot see the fault in my resolution and other methods I've tried give wrong answers as well.

Can anyone see my mistake/s ? Is it logical or arithmetical ?