Question

How high does a rocket have to go above Earth's surface before its weight is half of what it is on Earth?

Answer 1

Answer:

$h=1.6\times 10^6\ m$

Explanation:

As we know that the acceleration due to gravity decreases with height.

At certain height it will get to the half of its value on the surface of the earth.

As we know that the weight on the surface of the earth is given as:

$w=m.g$

where:

m = mass of the object

g = acceleration due to gravity of the substance

Since mass of the substance is constant so the variation is weight is possible only due to change in the acceleration due to gravity.

We know that the variation of the acceleration due gravity with height is given as:

$g_{_h}=g\times (1-\frac{2h}{R} )$

where:

$g_{_h}=$ value to acceleration due to gravity at height h

g = acceleration due to gravity at the earth's surface

h = height of the object

R = radius of the earth = $6400\ km$

according to question the weight becomes half, so,:

$4.9=9.8\times (1-\frac{2h}{6400\times10^3} )$

$h=1.6\times 10^6\ m$ is the height a rocket has to go above Earth's surface before its weight is half of what it is on Earth.

Answer 2

The height of the object must be the  $\bold {h = 1.6x 10^6}$ m in order to get the half of the weight, relative to the earth surface.

 

The weight on the earth surface is:

w = mg    

where,  

m = mass of the object  

g =  gravitational acceleration

 

Since, mass of the object is constant. thus the variation in weight is only due to change in gravitational acceleration.  

The variation of the acceleration due gravity with height,

$\bold {g_h = g (1- \dfrac {2h}{R})}$

 

where:  

gh -  gravitational acceleration at height h = 9.8/2 = 4.9 m/s  

g = gravitational acceleration at the earth's surface = 9.8 m/s  

h = height of the object = ?  

R = radius of the earth = 6400 km

Put the values in the equation we get,

$\bold {h = 1.6x 10^6}$ m

Therefore, the height of the object must be the  $\bold {h = 1.6x 10^6}$ m in order to get the half of the weight, relative to the earth surface.

To know more about  gravitational acceleration,

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