Question

A sattelite is at a distance of three time Earth's radius. What should be tte minimum speed given to it to escape Earth's gravitation, if Earth's radius is `6.4xx10^6` meters and accelaration due to gravity is `9.8` `ms^(-2)`?

Answer 1

Minimum speed should be `6.467` kilmeters per second.

Explanation:

For an object to escape from Earth, the kinetic energy given to it to must be equal to the work done against gravity going from earth to infinity.

As the force that we will be working on is gravity, the work done can be calculated uusing integral

`W=int_(3R)^oodW=int_(3R)^oo(GMm)/x^2dx`

= `GMm int_(3R)^oo1/x^2dx`

= `-GMm[1/x]_(3R)^oo`

= `GMm[1/(3R)-1/oo]=(GMm)/(3R)`

and if velocity provided is `v_e`, K.E. imparted to object would be `1/2mv_e^2` to escape from earth, we should have

`1/2mv_e^2=(GMm)/(3R)`

or `v_e=sqrt((2GM)/(3R)`

Note that accelaration due to gravity at earth's surface `g=(GM)/R^2` and hence `v_e=sqrt((2gR)/3)`

As `g=9.8m/s^2` and `R=6.4xx10^6m`

Escape velocity would be `v_e=sqrt((2xx9.8xx6.4xx10^6)/3)`

= `6.467xx10^3m/s` or `6.467(km)/s`