Question

find the value of acceleration due to gravity at a height equal to radius of earth from the surfaces of earth if the same value on earth surface is 9.8m/s².

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Answer 1

ANSWER

g∝R2GM

So, gp=(MeMp×(RpRe)2)g

=2×(21)2g

=2g=4.9 m/s2

yeah

Answer 2

Answer:

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Explanation:

Let:

Mass of the earth = M

Radius of the earth = R

Universal constant = G

Mass of a certain object = m

From Newton's law of universal gravitation.

When the object is on earth's surface, the gravie force exerted on it is given by

F = GMm/R^2________________(1)

The weight of the object on earth surface is given by

W = mg = 9.8*m___________(2)

Since the weight of a body is a measure of the gravitational influence on it. We can then equate (1) and (2)

GMm/R^2 = 9.8*m

GM = 9.8R^2____________(3)

Now let's consider when the object is at a height, h, above the earth's surface. The Newton's law will look like this

F' = GMm/(R + h)^2_________(4)

Let weight and gravity at the given height be W' and g'. Hence weight at this point is given as

W' = m*g'______________(5)

Again F' = W'

GMm/(R +h)^2 = m*g'

GM = g'(R + h)^2___________(6)

Equate (3) and (6)

9.8R^2 = g'(R + h)^2

g' = 9.8R^2/(R +h)^2___________(7)

In your question uou said the object is at a height equal radius of the earth. I want to assume you mean height from the earth's surface. If my interpretation is correct. It mean h = R. Substitute this in (7).

g' = 9.8R^2(R + R)^2

g' = 9.8*(R/2R)^2

g' = 9.8*0.25

g' = 2.45 m/s2.

Gravity at the required height is 2.45 m/s2.