Question

A body weighs 25kg on the surface of the Earth. if the mass of the earth is 6×10²⁴ kg ,the radius of
the earth is 6.4×10⁶m and the gravitational constant 6.7×10-¹¹Nm²kg-².
Calculate1) Mutual force of attraction between the body and the earth.
2) The acceleration produced in the body.
3) The acceleration produced in the earth

Answer 1

Answer:

Answer:

h = 4004.349 km

Explanation:

Given,

The mass of the Earth, M = 6 x 10²⁷ km

The radius of the Earth, R = 6.4 x 10³ km

At what height the acceleration of the meteor is, h = 4 m/s²

The acceleration due to gravity at the surface of the Earth is given by the formula,

g = GM/r²

At height 'h' acceleration is given by the formula,

g_{h}g

h

= GM/R_{h}^{2}R

h

2

Where,

R_{h}^{2}R

h

2

= R + h

R_{h}^{2}R

h

2

= GM/g_{h}g

h

R_{h}^{2}R

h

2

= 6.673 x 10⁻¹¹ X 6 x 10²⁴ / 4

= 1.00095 x 10¹⁴

R_{h}R

h

= 10004748.87 m

= 10004.749 km

From above,

R_{h}R

h

= R + h

h = R_{h}R

h

- R

= 10004.749 km - 6400 km

= 4004.349 km

Hence, the hieght of the meteor where its acceleration due to gravity is 4 m/s² is, h = 4004.349 km

Answer 2

$1) 1.5487 \times {10}^{ - 10}$

Explanation:

Explanation is given in above photo