المحاضرة (13)

Total Energy for circular orbital motion

 

When a body of mass m moving with speed v in circular orbit around another body of mass M where M>>m as the earth around the sun or satellite around the earth, the body of mass M is at rest with respect to the frame of reference.  The total energy of the two body system is the sum of the kinetic energy and the potential energy.

  E = K + U

*

From Newton’s second law  F = ma where a is the is the radial acceleration therefore,

     **

Multiply both sides by r/2

Substitute from equation ** into equation *, we get

The total energy for circular orbit

Note that the total energy is negative in a circular orbit.  And the kinetic energy is positive and equal to one half the magnitude of the potential energy.  The total energy called the binding energy for the system.

Escape velocity

باستخدام مفهوم الطاقة الكلية سنقوم بحساب سرعة الإفلات escape velocity من الجاذبية الأرضية.  وسرعة الإفلات هي أقل سرعة ابتدائية لجسم يقذف رأسياً ليتمكن الجسم من الإفلات من مجال الجاذبية الأرضية.

Suppose an object of mass m is projected vertically upward from the earth with initial speed v_i = v and r_i = R_e.  When the object is at maximum altitude, v_f = 0 and r_f = r_max. 

In this case the total energy of the system (Earth & object) is conserved, we can use the equation

solving for v_i² we get,

من هذه المعادلة إذا علمنا قيمة السرعة الابتدائية لانطلاق الجسم v_i يمكن حساب أقصى ارتفاع يمكن أن يصل إليه الجسم h حيث أن h = r_max-R_e.

لحساب سرعة الإفلات للجسم من مجال الجاذبية الأرضية مثل ما هو الحال عند إطلاق صاروخ فضائي أو مكوك من سطح الأرض إلى الفضاء الخارجي فإن سرعة الانطلاق الابتدائية التي يجب أن ينطلق بها المكوك يجب أن لا تقل عن سرعة الإفلات وإلا فإن المكوك سوف لن يصل إلى هدفه نتيجة لتأثير قوة الجاذبية. ولإيجاد سرعة الإفلات المطلوبة فإن ......

For the escape velocity the object will reach a final speed of v_f = 0 when r_max= ¥, therefore we substitute for v_i=v_esc and we get

Note that the escape velocity does not depends on the mass of the object projected from the earth.

This equation can be used to evaluate the escape velocity from any planet in the universe if the mass and the radius of the planet are known.

Escape velocities for the planets

Example

(a) Calculate the minimum energy required to send a 3000kg spacecraft from the earth to a distance point in space where earth’s gravity is negligible.  (b) If the journey is to take three weeks, what average power would the engine have to supply?

Solution

Example

A spaceship is fired from the Earth’s surface with an initial speed of 2×10⁴ m/s.  What will its speed when it is very far from the Earth? (Neglect friction.)

Solution

Energy is conserved between the surface and the distant point

       (K+U_g)_i + W_nc = (K+U_g)_f

Example

Two planets of masses m₁ and m₂ and radii r₁ and r₂, respectively, are nearly at rest when they are an infinite distance apart.  Because of their gravitational attraction, they head toward each other on a collision course. (a) When their center-to-center separation is d, find expressions for the speed of each planet and their relative velocity. (b) Find the kinetic energy of each planet just before they collide, if m₁=2×10²⁴ kg, m₂=8×10²⁴ kg, r₁=3×10⁶ m, and r₂=2×10⁶ m

Solution

(a)        At infinite separation, U=0; and at rest, K=0. Since energy is conserved, we have

The initial momentum is zero and momentum is conserved. Therefore

Combine equations (1) and (2) to find v₁ and v₂

The relative velocity is

(b)   Substitute for the given value for v₁ and v₂ we find that

v₁ = 1.03×10⁴ m/s and v₂= 2.58×10³ m/s.

Therefore,

and

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