Question

what is magnitude of the gravitational force between the earth and a 1kg object on its surface (mass of the earth is 6*10^24kg and radius of the Earth is 6.4 *10^6​

Answer 1

Answer:

F=GMm/r^2

F=(6.67×10^-11)(6×10^24)(1)/(6.4×10^6)^2

F=40.02×10^13/40.96×10^12

F=0.977×10

or approx

F=10 newton

hope it works

plz mark it as brainliest answer

Answer 2

Answer:

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by F =gm1m2/r²

Here,

m1 = mass of Earth = 6.0 x 1024kg

m2 = mass of the body = 1kg

r = distance between the two bodies

Radius of Earth = 6.4 x 106m

G = Universal gravitational constant = 6.67 x 10-11 Nm2kg-2

By substituting all the values in the equation

F = 6.67 x 1011 (6 x 1024 x 1 ) / (6.4 x 106)2

F = 9.8 N

This shows that Earth exerts a force of 9.8 N on a body of mass 1 kg. The body will exert an equal force of attraction of 9.8 N on the Earth.