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olasank [31]
3 years ago
6

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

speed of the object just before it hits Earth isgiven by: A. √GM/R B. √GM/2R C. √2GM/R D. √GM/R2 E. √GM/2R2
Physics
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

P.E=-\dfrac{GmM}{R+h}

The kinetic energy at height above the surface of the earth

K.E = 0

The total energy at height above the surface of the earth

E = K.E+P.E

E = -\dfrac{GmM}{R+h}....(I)

The total energy at the surface of the earth

E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}....(II)

We need to calculate the speed of the object  just before it hits Earth

From equation (I) and (II)

-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

Here, h = R

-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}

v= \sqrt{\dfrac{GM}{R}}

Hence, The speed of the object just before it hits Earth is  \sqrt{\dfrac{GM}{R}}.

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