PHYSICS MIDDLE SCHOOL

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earth's surface?

Answers

Answer 1

Answer:

Answer:

g = 4.7 × $10^(-16)$ m/ $s^(2)$

Explanation:

Given that the mass of the satellite = 700 kg, and 10,000 m above the earth;s surface.

From Newton's second law,

F = mg ............... 1

From Newton's gravitation law,

F = $(GMm)/(r^(2) )$ .................. 2

Where: F is the force, G is the gravitational constant, M is the mass of the first body, m is the mass of the second body, g is the gravitational force and r is the distance between the centers of the two bodies.

Equate 1 and 2 to have,

mg = $(GMm)/(r^(2) )$

⇒ g = $(GM)/(r^(2) )$

But; G = 6.67 × $10^(-11)$ N $m^(2) Kg^(-2)$ , M = 700 Kg, r = 10000 m

Thus,

g = $(6.67*10^(-11*700) )/(10000^(2) )$

= $(4.669*10^(-8) )/(1*10^(8) )$

= 4.669 × $10^(-16)$ m/ $s^(2)$

The force of gravity on the satellite is 4.7 × $10^(-16)$ m/ $s^(2)$ .

Answer 2

Answer:

The force of gravity on a 700kg satellite 10km above the Earth's surface can be calculated using the equation F = mg.

The force of gravity on an object is given by the equation F = mg, where F is the force of gravity, m is the mass of the object, and g is the acceleration due to gravity.

On Earth's surface, the acceleration due to gravity is approximately 9.8 m/s^2. However, as the satellite is 10 km above the Earth's surface, the force of gravity will be slightly less than that.

To calculate the force of gravity on the satellite, we can use the equation F = mg. Substituting the mass of the satellite (700 kg) and the acceleration due to gravity (9.8 m/s^2 - (9.8 m/s^2 * (10 km/ 6371 km))), we can find the force of gravity.

Learn more about force of gravity here:

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