When you drop a 0.40kg apple, earth exerts a force on it that accelerates it at 9.8m/s squared toward earths surface. According to newtons third law, the apple must exert an equal and opposite force on earth. If the mass of earth is 5.98 times 10 to the power of 24 kg, what is the magnitude of earths acceleration?

The answer is supposed to be 6.6 times 10 to the power of -25m/s squared.

Thank you very much to whoever can answer this! :)

Well, use Newton's first law to solve it, together with his third law.

Make use of F=ma

The first force (on apple) is given by F_a = (0.4)(9.8)

The second force (on Earth) is given by F_E = (5.98\times10^{24})(a)

Using the third law, F_a = F_E

You therefore have: (0.4)(9.8) = (5.98\times10^{24})(a)

Solve for a, the acceleration of the Earth! :)

I hope it helped! :)